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Calculus 1 : How to find differential functions

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Example Questions

Example Question #1342 : Calculus

Find the derivative of $\small x\small 2^{e^{x}}^{^{}}$ .

Possible Answers:

$2^{e^{x}}$

$\small e^{x}+x2^{e^{x}}$

$\small 2^{e^{x}}(xe^{x}ln2+1)$

$\small 2^{e^{x}}(xe^{x}ln2)$

$\small xe^{x}(2^{e^{x-1}})$

Correct answer:

$\small 2^{e^{x}}(xe^{x}ln2+1)$

Explanation:

The $\small 2^{e^{x}}$ is in the form a^u , in which $\small a$ is a constant and $\small u$ is a function of $\small x$ . This has the derivative (with respect to $\small x$ ) of,

(a^u)(du/dx)(lna) .

In this problem, $\small a=2$ and $\small u=e^x$ , so the derivative is,

$2^{e^{x}}(e^{x})(ln2)$ .

Since the $\small 2^{e^{x}}$ is being multiplied by , we can use the product rule to compute the entire derivative.

The derivative of $\small x$ is $\small 1$ , so using the product rule, we get the derivative to be $\small 2^{e^{x}}(xe^{x}ln2+1)$ .

Product rule:

$\small \frac{\mathrm{d} }{\mathrm{d} x}(u(x)v(x))=u(x)v'(x)+u'(x)v(x)$ .

In this case

$\small u(x)=x,u'(x)=1, v(x)=2^{e^{x}},v'(x)=2^{e^{x}}e^{x}ln2$

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Example Question #1343 : Calculus

Find the derivative of $\small \small ln(1/x)sinx$ .

Possible Answers:

$\small \small -cosx/x^2$

$\small -\small sinx/x$

$\small ln(1/x)cosx-sinx/x$

$\small \small cosx/x^2$

$\small \small -cosx/(x^2*lnx)$

Correct answer:

$\small ln(1/x)cosx-sinx/x$

Explanation:

The function $\small ln(1/x)$ is in the form $\small lnu$ , where $\small u$ is a function of $\small x$ .

The derivative of this is $\small u'/u$ .

In this case $\small \small u=1/x$ and $\small u'=-1/x^2$ , so the derivative is

$\small (-1/x^2)/(1/x)=(-1/x^2)\cdot (x)=-1/x$ .

The derivative of $\small sinx$ is $\small cosx$ .

Now we can use the product rule to get the total derivative.

$\small \small sinx\cdot(-1/x)+cosxln(1/x)=ln(1/x)cosx-sinx/x$

Product rule:

$\small \frac{\mathrm{d} }{\mathrm{d} x}(u(x)v(x))=u(x)v'(x)+u'(x)v(x)$ .

In this case,

$\small u(x)=ln(1/x),u'(x)=-1/x, v(x)=sinx, v'(x)=cosx$ .

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Example Question #131 : How To Find Differential Functions

Differentiate the function:

f(x)=2^4^0

Possible Answers:

f'(x)=40

f'(x)=x^4^0

$f'(x)=40\cdot 2^3^9$

f'(x)=0

Correct answer:

f'(x)=0

Explanation:

Becuase 2^4^0 is a constant, every derivative of will be 0.

The general rule for differentiating a constant , is as follows.

$f(x)=c\rightarrow f'(x)=0$

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Example Question #1345 : Calculus

Differentiate the function:

$f(x)=3-\frac{2}{3}x$

Possible Answers:

$f'(x)=1-\frac{2}{3}x$

$f'(x)=-\frac{2}{3}x^-^1$

f'(x)=3-x

$f'(x)=-\frac{2}{3}$

Correct answer:

$f'(x)=-\frac{2}{3}$

Explanation:

When the function has a sum or difference of terms, take the derivate with respect to the variable (in the case x) of each term:

$f(x)=3-\frac{2}{3}x$

Using the power rule which states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

we find our derivative to be,

$f'(x)=\frac{d}{dx}(3)-\frac{d}{dx}\left(\frac{2}{3}x\right)=0-1\cdot\frac{2}{3}\cdot x^1^-^1=-\frac{2}{3}$ .

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Example Question #132 : How To Find Differential Functions

Differentiate the equation:

$f(x)=\pi^2$

Possible Answers:

$f'(x)=2\pi$

f'(x)=0

{}f'(x)=2x

$f'(x)=2x\pi$

Correct answer:

f'(x)=0

Explanation:

Becuase $\pi^2$ is a constant, every derivative of will be 0.

The general rule for differentiating a constant , is as follows.

$f(x)=c\rightarrow f'(x)=0$

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Example Question #322 : Functions

Differentiate the function:

$f(t)=\frac{3}{4}t^1^2$

Possible Answers:

$f'(t)=\frac{3}{4}x^1^1$

f'(t)=9x^1^1

$f'(t)=\frac{3}{4}^8$

f'(t)=9x^1^2

Correct answer:

f'(t)=9x^1^1

Explanation:

Use the power rule: nx^n^-^1 and multiply the exponent by the coefficient then decrease the exponent by one to find the derivative of the function.

$f(t)=\frac{3}{4}t^1^2$

where n=12

$f'(t)=12\cdot \frac{3}{4}x^1^2^-^1=9x^1^1$

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Example Question #133 : How To Find Differential Functions

Differentiate the function:

f(a)=a^4-a+7

Possible Answers:

f'(a)=4a^4-a

f'(a)=a^3-1

f'(a)=4a^3-1

f'(a)=4a+6

Correct answer:

f'(a)=4a^3-1

Explanation:

When the function consists of the sum or differenct of terms, take the derivative of each term with respect to the variable (a).

Using the power rule which states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

we find our derivative to be,

$f'(a)=\frac{d}{da}(a^4)-\frac{d}{da}(a)+\frac{d}{da}(7)=4\cdot a^4^-^3-1\cdota^1^-^1+0=4a^3-1$ .

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Example Question #324 : Differential Functions

Differentiate the function:

f(x)=1.5x^2-3.4x+7

Possible Answers:

f'(x)=1.5x+3.4

f'(x)=3x-3.4

f'(x)=3x

f'(x)=5.1

Correct answer:

f'(x)=3x-3.4

Explanation:

When a function has a sum or difference of terms, take the derivative of each term with respect to x.

To take the derivative we will need to use the power rule which states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Applying this rule term by term, we find the derivative as follows.

$f'(x)=\frac{d}{dx}\left(1.5x^2\right)-\frac{d}{dx}(3.4x)+\frac{d}{dx}(7)$

$f'(x)=2\cdot 1.5x^2^-^1-1\cdot 3.4x^1^-^1+0$

f'(x)=3x-3.4

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Example Question #325 : Differential Functions

Differentiate the function:

f(b)=sin(b)-9cos(b)

Possible Answers:

f'(b)=cos(b)-9

f'(b)=-cos(b)-9sin(b)

f'(b)=-9sin(b)+cos(b)

f'(b)=cos(b)+9sin(b)

Correct answer:

f'(b)=cos(b)+9sin(b)

Explanation:

Take the derivative of each term with respect to b.

To take the derivative we will need to use the power rule which states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Also recall that the derivative of sin(x) is cos(x) .

Applying these rules we find the derivative to be:

$f'(b)=\frac{d}{dx}(sin(b))-\frac{d}{dx}(9cos(b))=cos(b)-9(-sin(b))=cos(b)+9sin(b)$ .

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Example Question #326 : Differential Functions

Differentiate the function:

$f(x)=4x^2-\cos(x)$

Possible Answers:

f'(x)=8x^2+sin(x)

f'(x)=4x^2-sin(x)

$f'(x)=8x+\sin(x)$

$f'(x)=4x-\sin(x)$

Correct answer:

$f'(x)=8x+\sin(x)$

Explanation:

When a function has a sum or difference of terms, take the derivative of each term with repspect to x.

To take the derivative we will need to use the power rule which states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Also recall that the derivative of cos(x) is -sin(x) .

Applying these rules, we find the derivative as follows.

$f'(x)=\frac{d}{dx}(4x^2)-\frac{d}{dx}(cos(x))=2\cdot 4x^2^-^1-(-sin(x))=8x+sin(x)$

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Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1342 : Calculus

Example Question #1343 : Calculus

Example Question #131 : How To Find Differential Functions

Example Question #1345 : Calculus

Example Question #132 : How To Find Differential Functions

Example Question #322 : Functions

Example Question #133 : How To Find Differential Functions

Example Question #324 : Differential Functions

Example Question #325 : Differential Functions

Example Question #326 : Differential Functions

All Calculus 1 Resources

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