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Calculus 1 : Differential Functions

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Example Questions

Example Question #311 : Differential Functions

Calculate the differential for the following function.

$y=\frac{x^2}{2}$

Possible Answers:

dy=x^2dx

dy=(2x)dx

$dy=\frac{2x}{x^2}dx$

dy=xdx

dy=(4x)dx

Correct answer:

dy=xdx

Explanation:

To solve this problem, you may use the quotient rule for finding derivatives. The quotient rule stipulates that for a function

$y=\frac{a}{b}$ , $y'=\frac{ba'-ab'}{b^2}$ .

In this problem, a=x^2 and b=2 .

Thus,

$\frac{dy}{dx}=\frac{2(2x)-x^2(0)}{4}$

$\frac{dy}{dx}=\frac{4x}{4}$

$\frac{dy}{dx}=x$ .

Therefore,

$\frac{dy}{dx}=x$ and dy=xdx .

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Example Question #121 : Other Differential Functions

Calculate the differential of the following function.

$y=\frac{1}{2}g^2+\frac{3}{5}g^3+\frac{2}{3}g$

Possible Answers:

$dy=\left(g+\frac{9}{5}g^2\right)dg$

$dy=\left(g^2+\frac{9}{5}g^2+\frac{2}{3}\right)dg$

$dy=\left(2g+\frac{9}{5}g^2+\frac{2}{3}\right)dg$

$dy=\left(g+\frac{9}{5}g^2+\frac{2}{5}\right)dg$

$dy=\left(g+\frac{9}{5}g^2+\frac{2}{3}\right)dg$

Correct answer:

$dy=\left(g+\frac{9}{5}g^2+\frac{2}{3}\right)dg$

Explanation:

Using the power rule, we can find the derivative of each part of the function. When using the power rule you multiply the coefficient by the exponent then decrease the exponent by one.

The derivative of $\frac{1}{2}g^2$ is $\frac{1}{2}\cdot 2g^{2-1}=g$ .

The derivative of $\frac{3}{5}g^3$ is $\frac{9}{5}g^2$ .

The derivative of $\frac{2}{3}g$ is $\frac{2}{3}$ .

When these derivatives are added together,

$\frac{dy}{dg}=g+\frac{9}{5}g^2+\frac{2}{3}$ .

Thus,

$dy=\left(g+\frac{9}{5}g^2+\frac{2}{3}\right)dg$

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Example Question #312 : Differential Functions

Calculate the differential for the following function.

$y=\frac{2x}{x-1}$

Possible Answers:

$dy=({(x-1)^2})dx$

$dy=\left(-\frac{2}{(x-1)^2}\right)dx$

$dy=\left(\frac{2}{(x-1)^2}\right)dx$

$dy=\left(-\frac{2x}{(x-1)^2}\right)dx$

$dy=2x-1 \ dx$

Correct answer:

$dy=\left(-\frac{2}{(x-1)^2}\right)dx$

Explanation:

Use the quotient rule to find this answer. The quotient rule dictates that for a function

$y=\frac{a}{b}$ , $y'=\frac{b(a')-a(b')}{b^2}$ .

For this particular question, a=2x and b=x-1 .

Apply the quotient rule to this function:

$\frac{dy}{dx}=\frac{2x-2-2x}{(x-1)^2}=-\frac{2}{(x-1)^2}$

$dy=\left(-\frac{2}{(x-1)^2}\right)dx$

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Example Question #125 : Other Differential Functions

Calculate the differential for the following function.

$y=\sqrt{x}$

Possible Answers:

$dy=x^{\frac{1}{2}} dx$

$dy=4\sqrt{x} dx$

$dy=\left(\frac{1}{2\sqrt{x}}\right)dx$

$dy=\left(\frac{1}{\sqrt{x}}\right)dx$

$dy=\left(\frac{2}{\sqrt{x}}\right)dx$

Correct answer:

$dy=\left(\frac{1}{2\sqrt{x}}\right)dx$

Explanation:

For finding the derivative of a root, it is helpful to turn the root into a power.

For example, in this problem it is helpful to turn the $\sqrt{x}$ into $x^\frac{1}{2}$ .

Now, we can easily apply the power rule,

$y=x^n\rightarrow dy=nx^{n-1}dx$

which yields the answer

$dy=\left(\frac{1}{2\sqrt{x}}\right)dx$ .

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Example Question #122 : Other Differential Functions

Calculate the differential for the following function.

$y=12x^{2}-\sqrt{x}$

Possible Answers:

dy=24x+12x^2 dx

$dy=\left(24x-\frac{1}{2\sqrt{x}}\right)dx$

$dy=\left(\frac{1}{2\sqrt{x}}\right)dx$

$dy=\left(24x^2-\frac{1}{2\sqrt{x}}\right)dx$

$dy=\left(24x-\frac{1}{\sqrt{x}}\right)dx$

Correct answer:

$dy=\left(24x-\frac{1}{2\sqrt{x}}\right)dx$

Explanation:

Using the power rule, we can solve this problem. The power rule states to multiply the coefficient with the exponent of the term then decrease the exponent by one.

The derivative of 12x^2 is 24x .

The derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$ .

Thus,

$\frac{dy}{dx}=24x-\frac{1}{2\sqrt{x}}$ and

$dy=\left(24x-\frac{1}{2\sqrt{x}}\right)dx$

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Example Question #313 : Differential Functions

Calculate the differential for the following.

y=x^2+3x+4x^3

Possible Answers:

dy=(2x+12x^2)dx

$dy=(4x^3+2)\ dx$

dy=(2x^2+3+12x^2)dx

dy=(2x+3+12x^2)dx

dy=(2x+3x+12x^2)dx

Correct answer:

dy=(2x+3+12x^2)dx

Explanation:

Use the power rule to differentiate this function. The power rule states to multiply the coefficient by the exponent then decrease the exponent by one.

The derivative of x^2 is .

The derivative of is .

The derivative of 4x^3 is 12x^2 .

Thus,

$\frac{dy}{dx}=2x+3+12x^2$ and dy=(2x+3+12x^2)dx

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Example Question #1341 : Calculus

Calculate the differential for the following.

$y=\frac{6x^2}{2-x}$

Possible Answers:

$dy=\frac{24x-6x^2+3}{(2-x)^2}\ dx$

$dy=\frac{24x-6x^2}{(2-x)^2}\ dx$

$dy=\frac{24x-6x^2}{4-x^2}\ dx$

$dy=\frac{4x^3}{2-x^2}\ dx$

$dy=\frac{24x^2-6x^2}{(2-x)^2}\ dx$

Correct answer:

$dy=\frac{24x-6x^2}{(2-x)^2}\ dx$

Explanation:

Use the quotient rule to find the solution to this problem. The quotient rule stipulates that for a function

$y=\frac{a}{b}$ , $y'=\frac{b(a')-a(b')}{b^2}$

In this problem, a=6x^2 and b=2-x .

Apply the quotient rule:

$\frac{dy}{dx}=\frac{2-x(12x)-6x^2(-1)}{(2-x)^2}$

$\frac{dy}{dx}=\frac{24x-6x^2}{(2-x)^2}$

$dy=\left(\frac{24x-6x^2}{(2-x)^2}\right)dx$

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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 #311 : Differential Functions

Differentiate the function:

f(x)=2^4^0

Possible Answers:

f'(x)=x^4^0

f'(x)=40

f'(x)=0

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

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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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #311 : Differential Functions

Example Question #121 : Other Differential Functions

Example Question #312 : Differential Functions

Example Question #125 : Other Differential Functions

Example Question #122 : Other Differential Functions

Example Question #313 : Differential Functions

Example Question #1341 : Calculus

Example Question #1342 : Calculus

Example Question #1343 : Calculus

Example Question #311 : Differential Functions

All Calculus 1 Resources

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