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Derivatives of Exponential and Logarithmic Functions
Find the derivative of exponential functions.
Derivative of the Exponential Function
Just as when we found the derivatives of other functions, we can find the derivatives of exponential and logarithmic functions using formulas. As we develop these formulas, we need to make certain basic assumptions. The proofs that these assumptions hold are beyond the scope of this course.
First of all, we begin with the assumption that the function \(B(x)={b}^{x},b>0,\) is defined for every real number and is continuous. In previous courses, the values of exponential functions for all rational numbers were defined, beginning with the definition of \({b}^{n},\) where \(n\) is a positive integer, as the product of \(b\) multiplied by itself \(n\) times. Later, we defined \({b}^{0}=1,{b}^{\text{-}n}=\frac{1}{{b}^{n}},\) for a positive integer \(n,\) and \({b}^{s\text{/}t}=(\sqrt[t]{b}{)}^{s}\) for positive integers \(s\) and \(t.\) These definitions leave open the question of the value of \({b}^{r}\) where \(r\) is an arbitrary real number. By assuming the continuity of \(B(x)={b}^{x},b>0,\) we may interpret \({b}^{r}\) as \(\underset{x\to r}{\text{lim}}{b}^{x}\) where the values of \(x\) as we take the limit are rational. For example, we may view \({4}^{\pi }\) as the number satisfying
\[\begin{array}{l}{4}^{3}<{4}^{\pi }<{4}^{4},{4}^{3.1}<{4}^{\pi }<{4}^{3.2},{4}^{3.14}<{4}^{\pi }<{4}^{3.15}, \\ {4}^{3.141}<{4}^{\pi }<{4}^{3.142},{4}^{3.1415}<{4}^{\pi }<{4}^{3.1416}\text{,}\ldots .\end{array}\]As we see in the following table, \({4}^{\pi }\approx 77.88.\)
| \(x\) | \({4}^{x}\) | \(x\) | \({4}^{x}\) |
| \(3\) | 64 | \(3.141593\) | 77.8802710486 |
| \(3.1\) | 73.5166947198 | \(3.1416\) | 77.8810268071 |
| \(3.14\) | 77.7084726013 | \(3.142\) | 77.9242251944 |
| \(3.141\) | 77.8162741237 | \(3.15\) | 78.7932424541 |
| \(3.1415\) | 77.8702309526 | \(3.2\) | 84.4485062895 |
| \(3.14159\) | 77.8799471543 | \(4\) | 256 |
We also assume that for \(B(x)={b}^{x},b>0,\) the value \({B}^{'}(0)\) of the derivative exists. In this section, we show that by making this one additional assumption, it is possible to prove that the function \(B(x)\) is differentiable everywhere.
We make one final assumption: that there is a unique value of \(b>0\) for which \({B}^{'}(0)=1.\) We define \(e\) to be this unique value, as we did in Introduction to Functions and Graphs. provides graphs of the functions \(y={2}^{x},y={3}^{x},y={2.7}^{x},\) and \(y={2.8}^{x}.\) A visual estimate of the slopes of the tangent lines to these functions at 0 provides evidence that the value of e lies somewhere between 2.7 and 2.8. The function \(E(x)={e}^{x}\) is called the natural exponential function. Its inverse, \(L(x)={\text{log}}_{e}x=\text{ln}\ x\) is called the natural logarithmic function.
For a better estimate of \(e,\) we may construct a table of estimates of \({B}^{'}(0)\) for functions of the form \(B(x)={b}^{x}.\) Before doing this, recall that
Condensed: the full section is in OpenStax Calculus Volume 1.
Derivative of the Logarithmic Function
Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.
If \(y={\text{log}}_{b}x,\) then \({b}^{y}=x.\) It follows that \(\text{ln}\ ({b}^{y})=\text{ln}\ x.\) Thus \(y\ \text{ln}\ b=\text{ln}\ x.\) Solving for \(y,\) we have \(y=\frac{\text{ln}\ x}{\text{ln}\ b}.\) Differentiating and keeping in mind that \(\text{ln}\ b\) is a constant, we see that
\[\frac{dy}{dx}=\frac{1}{x\ \text{ln}\ b}.\]The derivative in now follows from the chain rule.
If \(y={b}^{x},\) then \(\text{ln}\ y=x\ \text{ln}\ b.\) Using implicit differentiation, again keeping in mind that \(\text{ln}\ b\) is constant, it follows that \(\frac{1}{y}\ \frac{dy}{dx}=\text{ln}\ b.\) Solving for \(\frac{dy}{dx}\) and substituting \(y={b}^{x},\) we see that
\[\frac{dy}{dx}=y\ \text{ln}\ b={b}^{x}\text{ln}\ b.\]The more general derivative () follows from the chain rule.
□
Example
Try it.
Find the derivative of \(h(x)=\frac{{3}^{x}}{{3}^{x}+2}.\)
Solution
Use the quotient rule and .
\[\begin{array}{lllll}{h}^{'}(x) & =\frac{{3}^{x}\ \text{ln}\ 3({3}^{x}+2)-{3}^{x}\ \text{ln}\ 3({3}^{x})}{{({3}^{x}+2)}^{2}} & & & \text{Apply the quotient rule.} \\ & =\frac{2\cdot {3}^{x}\ \text{ln}\ 3}{{({3}^{x}+2)}^{2}} & & & \text{Simplify.}\end{array}\]Example
Try it.
Find the slope of the line tangent to the graph of \(y={\text{log}}_{2}(3x+1)\) at \(x=1.\)
Solution
To find the slope, we must evaluate \(\frac{dy}{dx}\) at \(x=1.\) Using , we see that
\[\frac{dy}{dx}=\frac{3}{(3x+1)\ \text{ln}\ 2}.\]By evaluating the derivative at \(x=1,\) we see that the tangent line has slope
\[\frac{dy}{dx}|\begin{array}{l} \\ {}_{x=1}\end{array}=\frac{3}{4\ \text{ln}\ 2}=\frac{3}{\ \text{ln}\ 16}.\]Condensed: the full section is in OpenStax Calculus Volume 1.
Logarithmic Differentiation
At this point, we can take derivatives of functions of the form \(y={(g(x))}^{n}\) for certain values of \(n,\) as well as functions of the form \(y={b}^{g(x)},\) where \(b>0\) and \(b\ne 1.\) Unfortunately, we still do not know the derivatives of functions such as \(y={x}^{x}\) or \(y={x}^{\pi }.\) These functions require a technique called logarithmic differentiation, which allows us to differentiate any function of the form \(h(x)=g{(x)}^{f(x)}.\) It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of \(y=\frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}.\) We outline this technique in the following problem-solving strategy.
Example
Try it.
Find the derivative of \(y={(2{x}^{4}+1)}^{\text{tan}\ x}.\)
Solution
Use logarithmic differentiation to find this derivative.
\[\begin{array}{llllll}\text{ln}\ y & = & \text{ln}{(2{x}^{4}+1)}^{\text{tan}\ x} & & & \text{Step 1. Take the natural logarithm of both sides.} \\ \text{ln}\ y & = & \text{tan}\ x\ \text{ln}\ (2{x}^{4}+1) & & & \text{Step 2. Expand using properties of logarithms.} \\ \frac{1}{y}\ \frac{dy}{dx} & = & {\text{sec}}^{2}x\ \text{ln}\ (2{x}^{4}+1)+\frac{8{x}^{3}}{2{x}^{4}+1}\cdot \text{tan}\ x & & & \begin{array}{l}\text{Step 3. Differentiate both sides. Use the} \\ \text{product rule on the right.}\end{array} \\ \frac{dy}{dx} & = & y\cdot ({\text{sec}}^{2}x\ \text{ln}\ (2{x}^{4}+1)+\frac{8{x}^{3}}{2{x}^{4}+1}\cdot \text{tan}\ x) & & & \text{Step 4. Multiply by}\ y\ \text{on both sides.} \\ \frac{dy}{dx} & = & {(2{x}^{4}+1)}^{\text{tan}\ x}({\text{sec}}^{2}x\ \text{ln}\ (2{x}^{4}+1)+\frac{8{x}^{3}}{2{x}^{4}+1}\cdot \text{tan}\ x) & & & \text{Step 5. Substitute}\ y={(2{x}^{4}+1)}^{\text{tan}\ x}.\end{array}\]Condensed: the full section is in OpenStax Calculus Volume 1.
Key Concepts
- On the basis of the assumption that the exponential function \(y={b}^{x},b>0\) is continuous everywhere and differentiable at 0, this function is differentiable everywhere and there is a formula for its derivative.
- We can use a formula to find the derivative of \(y=\text{ln}\ x,\) and the relationship \({\text{log}}_{b}x=\frac{\text{ln}\ x}{\text{ln}\ b}\) allows us to extend our differentiation formulas to include logarithms with arbitrary bases.
- Logarithmic differentiation allows us to differentiate functions of the form \(y=g{(x)}^{f(x)}\) or very complex functions by taking the natural logarithm of both sides and exploiting the properties of logarithms before differentiating.
Key Equations
| Derivative of the natural exponential function | \(\frac{d}{dx}({e}^{g(x)})={e}^{g(x)}{g}^{'}(x)\) |
| Derivative of the natural logarithmic function | \(\frac{d}{dx}(\text{ln}\ g(x))=\frac{1}{g(x)}{g}^{'}(x)\) |
| Derivative of the general exponential function | \(\frac{d}{dx}({b}^{g(x)})={b}^{g(x)}{g}^{'}(x)\ \text{ln}\ b\) |
| Derivative of the general logarithmic function | \(\frac{d}{dx}({\text{log}}_{b}g(x))=\frac{{g}^{'}(x)}{g(x)\ \text{ln}\ b}\) |
Derivatives of Exponential and Logarithmic Functions
For the following exercises, find \({f}^{'}(x)\) for each function.
For the following exercises, use logarithmic differentiation to find \(\frac{dy}{dx}.\)
For the following exercises, use the population of New York City from 1790 to 1860, given in the following table.
| Years since 1790 | Population |
| 0 | 33,131 |
| 10 | 60,515 |
| 20 | 96,373 |
| 30 | 123,706 |
| 40 | 202,300 |
| 50 | 312,710 |
| 60 | 515,547 |
| 70 | 813,669 |
Tagad tu Neviens kalkulators nenoslīgst šo, bet tā gabali ir komplējami. Izmēģiniet vienu zemāk vai ievadiet paši savu.
Prakse (40)
Izmēģiniet katru uz papīra vispirms. Atklājiet atbildi, lai pārbaudītu; pārbaudītas ones var atvērt risinātājs par katru soli.
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Find the derivative of \(f(x)={e}^{\text{tan}(2x)}.\)
Atbildēt uz šo jautājumu
Using the derivative formula and the chain rule,
\[\begin{array}{ll}{f}^{'}(x) & ={e}^{\text{tan}\ (2x)}\frac{d}{dx}(\text{tan}\ (2x)) \\ & ={e}^{\text{tan}(2x)}{\text{sec}}^{2}(2x)\cdot 2.\end{array}\] -
Find the derivative of \(y=\frac{{e}^{{x}^{2}}}{x}.\)
Atbildēt uz šo jautājumu
Use the derivative of the natural exponential function, the quotient rule, and the chain rule.
\[\begin{array}{lllll}{y}^{'} & =\frac{({e}^{{x}^{2}}\cdot 2)x\cdot x-1\cdot {e}^{{x}^{2}}}{{x}^{2}} & & & \text{Apply the quotient rule.} \\ & =\frac{{e}^{{x}^{2}}(2{x}^{2}-1)}{{x}^{2}} & & & \text{Simplify.}\end{array}\] -
Find the derivative of \(h(x)=x{e}^{2x}.\)
Atbildēt uz šo jautājumu
\({h}^{'}(x)={e}^{2x}+2x{e}^{2x}\)
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A colony of mosquitoes has an initial population of 1000. After \(t\) days, the population is given by \(A(t)=1000{e}^{0.3t}.\) Show that the ratio of the rate of change of the population, \({A}^{'}(t),\) to the population, \(A(t)\) is constant.
Atbildēt uz šo jautājumu
First find \({A}^{'}(t).\) By using the chain rule, we have \({A}^{'}(t)=300{e}^{0.3t}.\) Thus, the ratio of the rate of change of the population to the population is given by
\[\frac{{A}^{'}(t)}{A(t)}=\frac{300{e}^{0.3t}}{1000{e}^{0.3t}}=0.3.\]The ratio of the rate of change of the population to the population is the constant 0.3.
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If \(A(t)=1000{e}^{0.3t}\) describes the mosquito population after \(t\) days, as in the preceding example, what is the rate of change of \(A(t)\) after 4 days?
Atbildēt uz šo jautājumu
996
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Find the derivative of \(f(x)=\text{ln}\ ({x}^{3}+3x-4).\)
Atbildēt uz šo jautājumu
Use directly.
\[\begin{array}{lllll}{f}^{'}(x) & =\frac{1}{{x}^{3}+3x-4}\cdot (3{x}^{2}+3) & & & \text{Use}\ g(x)={x}^{3}+3x-4\ \text{in}\ {h}^{'}(x)=\frac{1}{g(x)}{g}^{'}(x). \\ & =\frac{3{x}^{2}+3}{{x}^{3}+3x-4} & & & \text{Rewrite.}\end{array}\] -
Find the derivative of \(f(x)=\text{ln}\ (\frac{{x}^{2}\text{sin}\ x}{2x+1}).\)
Atbildēt uz šo jautājumu
At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.
\[\begin{array}{llllll}f(x) & = & \text{ln}\ (\frac{{x}^{2}\text{sin}\ x}{2x+1})=2\ \text{ln}\ x+\text{ln}\ (\text{sin}\ x)-\text{ln}\ (2x+1) & & & \text{Apply properties of logarithms.} \\ {f}^{'}(x) & = & \frac{2}{x}+\text{cot}\ x-\frac{2}{2x+1} & & & \text{Apply sum rule and}\ {h}^{'}(x)=\frac{1}{g(x)}{g}^{'}(x).\end{array}\] -
Differentiate: \(f(x)=\text{ln}{(3x+2)}^{5}.\)
Atbildēt uz šo jautājumu
\({f}^{'}(x)=\frac{15}{3x+2}\)
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Find the derivative of \(h(x)=\frac{{3}^{x}}{{3}^{x}+2}.\)
Atbildēt uz šo jautājumu
Use the quotient rule and .
\[\begin{array}{lllll}{h}^{'}(x) & =\frac{{3}^{x}\ \text{ln}\ 3({3}^{x}+2)-{3}^{x}\ \text{ln}\ 3({3}^{x})}{{({3}^{x}+2)}^{2}} & & & \text{Apply the quotient rule.} \\ & =\frac{2\cdot {3}^{x}\ \text{ln}\ 3}{{({3}^{x}+2)}^{2}} & & & \text{Simplify.}\end{array}\] -
Find the slope of the line tangent to the graph of \(y={\text{log}}_{2}(3x+1)\) at \(x=1.\)
Atbildēt uz šo jautājumu
To find the slope, we must evaluate \(\frac{dy}{dx}\) at \(x=1.\) Using , we see that
\[\frac{dy}{dx}=\frac{3}{(3x+1)\ \text{ln}\ 2}.\]By evaluating the derivative at \(x=1,\) we see that the tangent line has slope
\[\frac{dy}{dx}|\begin{array}{l} \\ {}_{x=1}\end{array}=\frac{3}{4\ \text{ln}\ 2}=\frac{3}{\ \text{ln}\ 16}.\] -
Find the slope for the line tangent to \(y={3}^{x}\) at \(x=2.\)
Atbildēt uz šo jautājumu
\(9\ \text{ln}\ (3)\)
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Find the derivative of \(y={(2{x}^{4}+1)}^{\text{tan}\ x}.\)
Atbildēt uz šo jautājumu
Use logarithmic differentiation to find this derivative.
\[\begin{array}{llllll}\text{ln}\ y & = & \text{ln}{(2{x}^{4}+1)}^{\text{tan}\ x} & & & \text{Step 1. Take the natural logarithm of both sides.} \\ \text{ln}\ y & = & \text{tan}\ x\ \text{ln}\ (2{x}^{4}+1) & & & \text{Step 2. Expand using properties of logarithms.} \\ \frac{1}{y}\ \frac{dy}{dx} & = & {\text{sec}}^{2}x\ \text{ln}\ (2{x}^{4}+1)+\frac{8{x}^{3}}{2{x}^{4}+1}\cdot \text{tan}\ x & & & \begin{array}{l}\text{Step 3. Differentiate both sides. Use the} \\ \text{product rule on the right.}\end{array} \\ \frac{dy}{dx} & = & y\cdot ({\text{sec}}^{2}x\ \text{ln}\ (2{x}^{4}+1)+\frac{8{x}^{3}}{2{x}^{4}+1}\cdot \text{tan}\ x) & & & \text{Step 4. Multiply by}\ y\ \text{on both sides.} \\ \frac{dy}{dx} & = & {(2{x}^{4}+1)}^{\text{tan}\ x}({\text{sec}}^{2}x\ \text{ln}\ (2{x}^{4}+1)+\frac{8{x}^{3}}{2{x}^{4}+1}\cdot \text{tan}\ x) & & & \text{Step 5. Substitute}\ y={(2{x}^{4}+1)}^{\text{tan}\ x}.\end{array}\] -
Find the derivative of \(y=\frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}.\)
Atbildēt uz šo jautājumu
This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.
\[\begin{array}{llllll}\text{ln}\ y & = & \text{ln}\ \frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x} & & & \text{Step 1. Take the natural logarithm of both sides.} \\ \text{ln}\ y & = & \text{ln}\ x+\frac{1}{2}\ \text{ln}\ (2x+1)-x\ \text{ln}\ e-3\ \text{ln}\ \text{sin}\ x & & & \text{Step 2. Expand using properties of logarithms.} \\ \frac{1}{y}\ \frac{dy}{dx} & = & \frac{1}{x}+\frac{1}{2x+1}-1-3\frac{\text{cos}\ x}{\text{sin}\ x} & & & \text{Step 3. Differentiate both sides.} \\ \frac{dy}{dx} & = & y(\frac{1}{x}+\frac{1}{2x+1}-1-3\ \text{cot}\ x) & & & \text{Step 4. Multiply by}\ y\ \text{on both sides.} \\ \frac{dy}{dx} & = & \frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}(\frac{1}{x}+\frac{1}{2x+1}-1-3\ \text{cot}\ x) & & & \text{Step 5. Substitute}\ y=\frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}.\end{array}\] -
Find the derivative of \(y={x}^{r}\) where \(r\) is an arbitrary real number.
Atbildēt uz šo jautājumu
The process is the same as in , though with fewer complications.
\[\begin{array}{llllll}\text{ln}\ y & = & \text{ln}{x}^{r} & & & \text{Step 1. Take the natural logarithm of both sides.} \\ \text{ln}\ y & = & r\ \text{ln}\ x & & & \text{Step 2. Expand using properties of logarithms.} \\ \frac{1}{y}\ \frac{dy}{dx} & = & r\frac{1}{x} & & & \text{Step 3. Differentiate both sides.} \\ \frac{dy}{dx} & = & y\frac{r}{x} & & & \text{Step 4. Multiply by}\ y\ \text{on both sides.} \\ \frac{dy}{dx} & = & {x}^{r}\frac{r}{x} & & & \text{Step 5. Substitute}\ y={x}^{r}. \\ \frac{dy}{dx} & = & r{x}^{r-1} & & & \text{Simplify.}\end{array}\] -
Use logarithmic differentiation to find the derivative of \(y={x}^{x}.\)
Atbildēt uz šo jautājumu
\(\frac{dy}{dx}={x}^{x}(1+\text{ln}\ x)\)
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Find the derivative of \(y={(\text{tan}\ x)}^{\pi }.\)
Atbildēt uz šo jautājumu
\({y}^{'}=\pi {(\text{tan}\ x)}^{\pi -1}{\text{sec}}^{2}x\)
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\(f(x)={x}^{2}{e}^{x}\)
Atbildēt uz šo jautājumu
\(2x{e}^{x}+{x}^{2}{e}^{x}\)
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\(f(x)=\frac{{e}^{\text{-}x}}{x}\)
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\(f(x)={e}^{{x}^{3}\text{ln}\ x}\)
Atbildēt uz šo jautājumu
\({e}^{{x}^{3}\text{ln}\ x}(3{x}^{2}\text{ln}\ x+{x}^{2})\)
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\(f(x)=\sqrt{{e}^{2x}+2x}\)
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\(f(x)=\frac{{e}^{x}-{e}^{\text{-}x}}{{e}^{x}+{e}^{\text{-}x}}\)
Atbildēt uz šo jautājumu
\(\frac{4}{{({e}^{x}+{e}^{\text{-}x})}^{2}}\)
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\(f(x)=\frac{{10}^{x}}{\text{ln}\ 10}\)
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\(f(x)={2}^{4x}+4{x}^{2}\)
Atbildēt uz šo jautājumu
\({2}^{4x+2}\cdot \text{ln}\ 2+8x\)
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\(f(x)={3}^{\text{sin}\ 3\text{x}}\)
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\(f(x)={x}^{\pi }\cdot {\pi }^{x}\)
Atbildēt uz šo jautājumu
\(\pi {x}^{\pi -1}\cdot {\pi }^{x}+{x}^{\pi }\cdot {\pi }^{x}\text{ln}\ \pi\)
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\(f(x)=\text{ln}\ (4{x}^{3}+x)\)
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\(f(x)=\text{ln}\sqrt{5x-7}\)
Atbildēt uz šo jautājumu
\(\frac{5}{2(5x-7)}\)
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\(f(x)={x}^{2}\text{ln}\ 9x\)
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\(f(x)=\text{log}\ (\text{sec}\ x)\)
Atbildēt uz šo jautājumu
\(\frac{\text{tan}\ x}{\text{ln}\ 10}\)
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\(f(x)={\text{log}}_{7}{(6{x}^{4}+3)}^{5}\)
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\(f(x)={2}^{x}\cdot {\text{log}}_{3}{7}^{{x}^{2}-4}\)
Atbildēt uz šo jautājumu
\({2}^{x}\cdot \text{ln}\ 2\cdot {\text{log}}_{3}{7}^{{x}^{2}-4}+{2}^{x}\cdot \frac{2x\ \text{ln}\ 7}{\text{ln}\ 3}\)
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\(y={x}^{\sqrt{x}}\)
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\(y={(\text{sin}\ 2x)}^{4x}\)
Atbildēt uz šo jautājumu
\({(\text{sin}\ 2x)}^{4x}[4\cdot \text{ln}\ (\text{sin}\ 2x)+8x\cdot \text{cot}\ 2x]\)
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\(y={(\text{ln}\ x)}^{\text{ln}\ x}\)
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\(y={x}^{{\text{log}}_{2}x}\)
Atbildēt uz šo jautājumu
\({x}^{{\text{log}}_{2}x}\cdot \frac{2\ \text{ln}\ x}{x\ \text{ln}\ 2}\)
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\(y={({x}^{2}-1)}^{\text{ln}\ x}\)
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\(y={x}^{\text{cot}\ x}\)
Atbildēt uz šo jautājumu
\({x}^{\text{cot}\ x}\cdot [\text{-}{\text{csc}}^{2}x\cdot \text{ln}\ x+\frac{\text{cot}\ x}{x}]\)
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\(y=\frac{x+11}{\sqrt[3]{{x}^{2}-4}}\)
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\(y={x}^{-1\text{/}2}{({x}^{2}+3)}^{2\text{/}3}{(3x-4)}^{4}\)
Atbildēt uz šo jautājumu
\({x}^{-1\text{/}2}{({x}^{2}+3)}^{2\text{/}3}{(3x-4)}^{4}\cdot [\frac{-1}{2x}+\frac{4x}{3({x}^{2}+3)}+\frac{12}{3x-4}]\)
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[T] Find an equation of the tangent line to the graph of \(f(x)=4x{e}^{({x}^{2}-1)}\) at the point where
\(x=-1.\) Graph both the function and the tangent line.
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Kā lietot: Derivatives of Exponential and Logarithmic Functions
- Find the derivative of exponential functions.
- Find the derivative of logarithmic functions.
- Use logarithmic differentiation to determine the derivative of a function.
- If,
- If
- To differentiate
- Use properties of logarithms to expand
- Differentiate both sides of the equation. On the left we will have
Jautājumi, ko cilvēki vaicā
What is a derivative in one sentence?
The slope of the graph at a point, the rate at which the output is changing there. Speed is the derivative of position.
What is an integral in one sentence?
The accumulated total of a rate: the area under the curve. Distance travelled is the integral of speed.
Why are derivatives and integrals opposites?
That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.
When do I use substitution and when integration by parts?
Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function, a polynomial times an exponential, log or trig function.
Daļas šīs lapas ir pielāgotas no OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Šeit ir pārliecināts un vēlreiz izskaidrots; kļūdas ir mūsu. Tā kā tās avots ir licencēts nekomerciālai izmantošanai, šī nodarbība ir bezmaksas, to nekas nepārdod un tiek dalīta ar to pašu licenci.
Vairāk Calculus
LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests