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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #391 : Partial Derivatives

Find $\frac{\partial f}{dx}$ $\displaystyle \frac{\partial f}{dx}$.

$f(x,y,z)=xyz^{10}+\frac{xy}{\ln(yz)}+10e^z+e^{xy}$ $\displaystyle f(x,y,z)=xyz^{10}+\frac{xy}{\ln(yz)}+10e^z+e^{xy}$

Possible Answers:

$\frac{\partial f}{dx}=xz^{10}+\frac{1}{\ln(yz)}+xe^{xy}$ $\displaystyle \frac{\partial f}{dx}=xz^{10}+\frac{1}{\ln(yz)}+xe^{xy}$

$\frac{\partial f}{dx}=xyz^{10}+\frac{xy}{\ln(yz)}+ye^{xy}$ $\displaystyle \frac{\partial f}{dx}=xyz^{10}+\frac{xy}{\ln(yz)}+ye^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}$ $\displaystyle \frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}$

$\frac{\partial f}{dx}=\frac{y}{\ln(yz)}+ye^{xy}$ $\displaystyle \frac{\partial f}{dx}=\frac{y}{\ln(yz)}+ye^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$ $\displaystyle \frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

Correct answer:

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$ $\displaystyle \frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

Explanation:

In order to find $\frac{\partial f}{dx}$ $\displaystyle \frac{\partial f}{dx}$, we need to take the derivative of $\displaystyle f$ in respect to $\displaystyle x$, and treat $\displaystyle y$, and $\displaystyle z$ as constants. We also need to remember what the derivatives of natural log, exponential functions and power functions are for single variables.

Natural Log:

$f(x)=\ln(g(x))$ $\displaystyle f(x)=\ln(g(x))$

$f'(x)=\frac{g'(x)}{g(x)}$ $\displaystyle f'(x)=\frac{g'(x)}{g(x)}$

Exponential Functions:

$f(x)=e^{g(x)}$ $\displaystyle f(x)=e^{g(x)}$

$f'(x)=g'(x)e^{g(x)}$ $\displaystyle f'(x)=g'(x)e^{g(x)}$

Power Functions:

$\displaystyle f(x)=x^n$

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

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+0+ye^{xy}$ $\displaystyle \frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+0+ye^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$ $\displaystyle \frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

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Example Question #392 : Partial Derivatives

True or False

$\frac{\partial}{\partial y}\Big(x^x+wz\ln(uvxw)\Big)=0$ $\displaystyle \frac{\partial}{\partial y}\Big(x^x+wz\ln(uvxw)\Big)=0$

Possible Answers:

False

True

Correct answer:

True

Explanation:

True:

Since there are no y's $\displaystyle y's$ in the equation, the derivative of a constant is $\displaystyle 0$.

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Example Question #391 : Partial Derivatives

What is the partial derivative $\displaystyle f_x(x,y)$ of the function

$f(x,y)=e^{xy}$ $\displaystyle f(x,y)=e^{xy}$?

Possible Answers:

$f_x(x,y)=ye^{xy}$ $\displaystyle f_x(x,y)=ye^{xy}$

$f_x(x,y)=xe^{xy}$ $\displaystyle f_x(x,y)=xe^{xy}$

$f_x(x,y)=xye^{xy}$ $\displaystyle f_x(x,y)=xye^{xy}$

$f_x(x,y)=e^{xy}$ $\displaystyle f_x(x,y)=e^{xy}$

Correct answer:

$f_x(x,y)=ye^{xy}$ $\displaystyle f_x(x,y)=ye^{xy}$

Explanation:

We can find $\displaystyle f_x(x,y)$ given $f(x,y)=e^{xy}$ $\displaystyle f(x,y)=e^{xy}$ by differentiating the function while holding $\displaystyle y$ constant, i.e. we treat $\displaystyle y$ as a number. So we get

$f_x(x,y)=\frac{\partial}{\partial x}e^{xy}=e^{xy}\frac{\partial}{\partial x}xy=e^{xy}\cdot y=ye^{xy}$ $\displaystyle f_x(x,y)=\frac{\partial}{\partial x}e^{xy}=e^{xy}\frac{\partial}{\partial x}xy=e^{xy}\cdot y=ye^{xy}$

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Example Question #392 : Partial Derivatives

Find the partial derivative $\displaystyle f_y(x,y)$ for the function $\displaystyle f(x,y)=2xy+2$.

Possible Answers:

$\displaystyle f_y(x,y)=2y$

$\displaystyle f_y(x,y)=0$

$\displaystyle f_y(x,y)=2xy$

$\displaystyle f_y(x,y)=2x$

Correct answer:

$\displaystyle f_y(x,y)=2x$

Explanation:

We can find $\displaystyle f_y(x,y)$ from the function $\displaystyle f(x,y)=2xy+2$ by taking the derivative holding $\displaystyle x$ constant and letting $\displaystyle y$ be the variable. This means

$f_y(x,y)=\frac{\partial}{\partial y}(2xy+2)=2x$ $\displaystyle f_y(x,y)=\frac{\partial}{\partial y}(2xy+2)=2x$

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Example Question #393 : Partial Derivatives

Given the function $f(x,y)=\sin x^2y$ $\displaystyle f(x,y)=\sin x^2y$, find the partial derivative $\displaystyle f_x(x,y)$.

Possible Answers:

$f_x(x,y)=2x\cos x^2y$ $\displaystyle f_x(x,y)=2x\cos x^2y$

$f_x(x,y)=y\cos x^2y$ $\displaystyle f_x(x,y)=y\cos x^2y$

$f_x(x,y)=2xy\cos x^2y$ $\displaystyle f_x(x,y)=2xy\cos x^2y$

$f_x(x,y)=x^2y\cos x^2y$ $\displaystyle f_x(x,y)=x^2y\cos x^2y$

Correct answer:

$f_x(x,y)=2xy\cos x^2y$ $\displaystyle f_x(x,y)=2xy\cos x^2y$

Explanation:

To find the partial derivative $\displaystyle f_x(x,y)$ of the function $f(x,y)=\sin x^2y$ $\displaystyle f(x,y)=\sin x^2y$, we calculate the derivative of f(x,y) $\displaystyle f(x,y)$ while treating $\displaystyle y$ as a constant. So we get

$f_x(x,y)=\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\sin x^2y=(\cos x^2y)\cdot \frac{\partial}{\partial x}x^2y$ $\displaystyle f_x(x,y)=\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\sin x^2y=(\cos x^2y)\cdot \frac{\partial}{\partial x}x^2y$

$=2xy\cos x^2y$ $\displaystyle =2xy\cos x^2y$

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Example Question #394 : Partial Derivatives

Given the function $f(x,y)=\sin x^2y$ $\displaystyle f(x,y)=\sin x^2y$, find the partial derivative $\displaystyle f_y(x,y)$.

Possible Answers:

$f_y(x,y)=2xy\cos x^2y$ $\displaystyle f_y(x,y)=2xy\cos x^2y$

$f_y(x,y)=y\cos x^2y$ $\displaystyle f_y(x,y)=y\cos x^2y$

$f_y(x,y)=2x\cos x^2y$ $\displaystyle f_y(x,y)=2x\cos x^2y$

$f_y(x,y)=y\cos x^2y$ $\displaystyle f_y(x,y)=y\cos x^2y$

$f_y(x,y)=x^2\cos x^2y$ $\displaystyle f_y(x,y)=x^2\cos x^2y$

$f_y(x,y)=2xy\cos x^2y$ $\displaystyle f_y(x,y)=2xy\cos x^2y$

Correct answer:

$f_y(x,y)=x^2\cos x^2y$ $\displaystyle f_y(x,y)=x^2\cos x^2y$

Explanation:

To find the partial derivative $\displaystyle f_y(x,y)$ of the function $f(x,y)=\sin x^2y$ $\displaystyle f(x,y)=\sin x^2y$, we calculate the derivative of f(x,y) $\displaystyle f(x,y)$ while treating $\displaystyle x$ as a constant. So we get

$f_y(x,y)=\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\sin x^2y=(\cos x^2y)\cdot \frac{\partial}{\partial y}x^2y$ $\displaystyle f_y(x,y)=\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\sin x^2y=(\cos x^2y)\cdot \frac{\partial}{\partial y}x^2y$

$=x^2\cos x^2y$ $\displaystyle =x^2\cos x^2y$

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Example Question #395 : Partial Derivatives

Given the function $f(x,y)=\ln (1+x^2+y^2)$ $\displaystyle f(x,y)=\ln (1+x^2+y^2)$, find the partial derivative $\displaystyle f_x(x,y)$.

Possible Answers:

$\displaystyle f_x(x,y)=0$

$f_x(x,y)=\frac{2x}{1+x^2+y^2}$ $\displaystyle f_x(x,y)=\frac{2x}{1+x^2+y^2}$

$f_x(x,y)=\frac{2y}{1+x^2+y^2}$ $\displaystyle f_x(x,y)=\frac{2y}{1+x^2+y^2}$

$f_x(x,y)=\frac{2x+2y}{1+x^2+y^2}$ $\displaystyle f_x(x,y)=\frac{2x+2y}{1+x^2+y^2}$

Correct answer:

$f_x(x,y)=\frac{2x}{1+x^2+y^2}$ $\displaystyle f_x(x,y)=\frac{2x}{1+x^2+y^2}$

Explanation:

To find the partial derivative $\displaystyle f_x(x,y)$ of the function $f(x,y)=\ln (1+x^2+y^2)$ $\displaystyle f(x,y)=\ln (1+x^2+y^2)$, we differentiate the function with respect to $\displaystyle x$, which means we hold $\displaystyle y$ constant. Then we get (using the chain rule):

$f_x(x,y)=\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\ln (1+x^2+y^2)=\frac{1}{1+x^2+y^2}\cdot \frac{\partial}{\partial x}(1+x^2+y^2)$ $\displaystyle f_x(x,y)=\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\ln (1+x^2+y^2)=\frac{1}{1+x^2+y^2}\cdot \frac{\partial}{\partial x}(1+x^2+y^2)$

$=\frac{2x}{1+x^2+y^2}$ $\displaystyle =\frac{2x}{1+x^2+y^2}$

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Example Question #396 : Partial Derivatives

Given the function $f(x,y)=\ln (2+x^2+y^2)$ $\displaystyle f(x,y)=\ln (2+x^2+y^2)$, find the partial derivative $\displaystyle f_y(x,y)$.

Possible Answers:

$\displaystyle f_y(x,y)=0$

$f_y(x,y)=\frac{2x}{2+x^2+y^2}$ $\displaystyle f_y(x,y)=\frac{2x}{2+x^2+y^2}$

$f_y(x,y)=\frac{2x+2y}{1+x^2+y^2}$ $\displaystyle f_y(x,y)=\frac{2x+2y}{1+x^2+y^2}$

$f_y(x,y)=\frac{2y}{1+x^2+y^2}$ $\displaystyle f_y(x,y)=\frac{2y}{1+x^2+y^2}$

Correct answer:

$f_y(x,y)=\frac{2y}{1+x^2+y^2}$ $\displaystyle f_y(x,y)=\frac{2y}{1+x^2+y^2}$

Explanation:

To find the partial derivative $\displaystyle f_y(x,y)$ of the function $f(x,y)=\ln (2+x^2+y^2)$ $\displaystyle f(x,y)=\ln (2+x^2+y^2)$, we differentiate the function with respect to $\displaystyle y$, which means we hold $\displaystyle x$ constant. Then we get (using the chain rule):

$f_y(x,y)=\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\ln (2+x^2+y^2)=\frac{1}{2+x^2+y^2}\cdot \frac{\partial}{\partial y}(2+x^2+y^2)$ $\displaystyle f_y(x,y)=\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\ln (2+x^2+y^2)=\frac{1}{2+x^2+y^2}\cdot \frac{\partial}{\partial y}(2+x^2+y^2)$

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

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Example Question #397 : Partial Derivatives

Given the function $f(x,y)=e^{24xy+25x^2}$ $\displaystyle f(x,y)=e^{24xy+25x^2}$, find the partial derivative

$\displaystyle f_x(x,y)$

Possible Answers:

$f_x(x,y)=(24y+50x)e^{24xy+25x^2}$ $\displaystyle f_x(x,y)=(24y+50x)e^{24xy+25x^2}$

$f_x(x,y)=24xe^{24xy+25x^2}\,$ $\displaystyle f_x(x,y)=24xe^{24xy+25x^2}\,$

$f_x(x,y)=e^{24y+50x}$ $\displaystyle f_x(x,y)=e^{24y+50x}$

$f_x(x,y)=(24y+50x)e^{24y+50x}$ $\displaystyle f_x(x,y)=(24y+50x)e^{24y+50x}$

Correct answer:

$f_x(x,y)=(24y+50x)e^{24xy+25x^2}$ $\displaystyle f_x(x,y)=(24y+50x)e^{24xy+25x^2}$

Explanation:

To find the partial derivative f_x $\displaystyle f_x$ of $f(x,y)=e^{24xy+25x^2}$ $\displaystyle f(x,y)=e^{24xy+25x^2}$, we need to differentiate f(x,y) $\displaystyle f(x,y)$ with respect to $\displaystyle x$ while holding $\displaystyle y$ constant. We can use the chain rule to get

$f_x(x,y)=\frac{\partial}{\partial x}e^{24xy+25x^2}=e^{24xy+25x^2}\cdot \frac{\partial}{\partial x}(24xy+25x^2)$ $\displaystyle f_x(x,y)=\frac{\partial}{\partial x}e^{24xy+25x^2}=e^{24xy+25x^2}\cdot \frac{\partial}{\partial x}(24xy+25x^2)$

$=(24y+50x)e^{24xy+25x^2}$ $\displaystyle =(24y+50x)e^{24xy+25x^2}$

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Example Question #398 : Partial Derivatives

Given the function $f(x,y)=e^{24xy+25x^2}$ $\displaystyle f(x,y)=e^{24xy+25x^2}$, find the partial derivative

$\displaystyle f_y(x,y)$

Possible Answers:

$f_y(x,y)=24xe^{24xy+25x^2}$ $\displaystyle f_y(x,y)=24xe^{24xy+25x^2}$

$f_y(x,y)=e^{24y+50x}$ $\displaystyle f_y(x,y)=e^{24y+50x}$

$f_y(x,y)=(24y+50x)e^{24xy+25x^2}\,$ $\displaystyle f_y(x,y)=(24y+50x)e^{24xy+25x^2}\,$

$f_y(x,y)=(24y+50x)e^{24y+50x}$ $\displaystyle f_y(x,y)=(24y+50x)e^{24y+50x}$

Correct answer:

$f_y(x,y)=24xe^{24xy+25x^2}$ $\displaystyle f_y(x,y)=24xe^{24xy+25x^2}$

Explanation:

To find the partial derivative f_y $\displaystyle f_y$ of $f(x,y)=e^{24xy+25x^2}$ $\displaystyle f(x,y)=e^{24xy+25x^2}$, we need to differentiate f(x,y) $\displaystyle f(x,y)$ with respect to $\displaystyle y$ while holding $\displaystyle x$ constant. We can use the chain rule to get

$f_y(x,y)=\frac{\partial}{\partial y}e^{24xy+25x^2}=e^{24xy+25x^2}\cdot \frac{\partial}{\partial y}(24xy+25x^2)$ $\displaystyle f_y(x,y)=\frac{\partial}{\partial y}e^{24xy+25x^2}=e^{24xy+25x^2}\cdot \frac{\partial}{\partial y}(24xy+25x^2)$

$=24xe^{24xy+25x^2}$ $\displaystyle =24xe^{24xy+25x^2}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #391 : Partial Derivatives

Example Question #392 : Partial Derivatives

Example Question #391 : Partial Derivatives

Example Question #392 : Partial Derivatives

Example Question #393 : Partial Derivatives

Example Question #394 : Partial Derivatives

Example Question #395 : Partial Derivatives

Example Question #396 : Partial Derivatives

Example Question #397 : Partial Derivatives

Example Question #398 : Partial Derivatives

All Calculus 3 Resources

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