Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #1041 : Partial Derivatives

Find \(\displaystyle f_{xx}\) of the following function:

\(\displaystyle f(x,y)=x^3+y\tan(xy)\)

Possible Answers:

\(\displaystyle 6x+y^2\sec(xy)\tan(xy)\)

\(\displaystyle 6x+2y^3\sec(xy)\tan(xy)\)

\(\displaystyle 6x+2y^2\sec(xy)\tan(xy)\)

\(\displaystyle 6x+y^3\sec(xy)\tan(xy)\)

\(\displaystyle 3x^2+2y^3\sec(xy)\tan(xy)\)

Correct answer:

\(\displaystyle 6x+2y^3\sec(xy)\tan(xy)\)

Explanation:

To find the given partial derivative of the function, we must treat the other variable(s) as constants. For higher order partial derivatives, we work from left to right for the given variables.

First, we must find the partial derivative of the function with respect to x:

\(\displaystyle f_x=3x^2+y^2\sec^2(xy)\)

Then, we find the partial derivative of this function with respect to x:

\(\displaystyle f_{xx}=6x+2y^3\sec(xy)\tan(xy)\)

Example Question #1042 : Partial Derivatives

Find \(\displaystyle f_{xxz}\) of the function \(\displaystyle x^2z^2+x^3\cos(z)\)

Possible Answers:

\(\displaystyle 4z+6x\sin(z)\)

\(\displaystyle 4z-x\cos(z)\)

\(\displaystyle 4z-6x\sin(z)\)

\(\displaystyle 4z^2-6x\sin(z)\)

Correct answer:

\(\displaystyle 4z-6x\sin(z)\)

Explanation:

To find \(\displaystyle f_{xxz}\) of the function, you take three consecutive partial derivatives:

\(\displaystyle \frac{\partial }{\partial x}(x^2z^2+x^3\cos(z))=2xz^2+3x^2\cos(z)\)

\(\displaystyle \frac{\partial }{\partial x}(2xz^2+3x^2\cos(z))=2z^2+6x\cos(z)\)

\(\displaystyle \frac{\partial }{\partial z}(2z^2+6x\cos(z))=4z-6x\sin(z)\)

Example Question #1043 : Partial Derivatives

Find \(\displaystyle f_{xxz}\) of the function \(\displaystyle 5x^2z^5+ze^{5x}\)

Possible Answers:

\(\displaystyle 50z^4-25e^{10x}\)

\(\displaystyle 50z^3+20e^{5x}\)

\(\displaystyle 50z^4+25e^{5x}\)

\(\displaystyle 50z^5+20e^{4x}\)

Correct answer:

\(\displaystyle 50z^4+25e^{5x}\)

Explanation:

To find \(\displaystyle f_{xxz}\) of the function, you must take three consecutive partial derivatives:

\(\displaystyle \frac{\partial }{\partial x}(5x^2z^5+ze^{5x})=10xz^5+5ze^{5x}\)

\(\displaystyle \frac{\partial }{\partial x}(10xz^5+5ze^{5x})=10z^5+25ze^{5x}\)

\(\displaystyle \frac{\partial }{\partial z}(10z^5+25ze^{5x})=50z^4+25e^{5x}\)

Example Question #1044 : Partial Derivatives

Find \(\displaystyle \frac{\partial F}{\partial y}\) given

\(\displaystyle F(x,y)=e^x+e^y+cos\,x+sin\,y\)

Possible Answers:

\(\displaystyle \frac{\partial F}{\partial y}=e^{xy}+cos\,y-sin\,x\)

\(\displaystyle \frac{\partial F}{\partial y}=e^x-sin\,x\)

\(\displaystyle \frac{\partial F}{\partial y}=e^y+cos\,y\)

Correct answer:

\(\displaystyle \frac{\partial F}{\partial y}=e^y+cos\,y\)

Explanation:

To find \(\displaystyle \frac{\partial F}{\partial y}\) we take the derivative of the function \(\displaystyle F(x,y)\) with respect to \(\displaystyle y\) and treat the other variables as constants. 

As such,

\(\displaystyle \frac{\partial }{\partial y}[F(x,y)]=\frac{\partial }{\partial y}[e^x+e^y+cos\,x+sin\,y]\)

yields

\(\displaystyle \frac{\partial F}{\partial y}=e^y+cos\,y\)

Example Question #1045 : Partial Derivatives

Find \(\displaystyle \frac{\partial F}{\partial x}\) given

\(\displaystyle F(x,y)=tan\,x-cot\,y+y^2\)

Possible Answers:

\(\displaystyle \frac{\partial F}{\partial x}=csc^2\,y+2y\)

\(\displaystyle \frac{\partial F}{\partial x}=0\)

\(\displaystyle \frac{\partial F}{\partial x}=sec^2\,x\)

Correct answer:

\(\displaystyle \frac{\partial F}{\partial x}=sec^2\,x\)

Explanation:

To find \(\displaystyle \frac{\partial F}{\partial x}\) we take the derivative of the function \(\displaystyle F(x,y)\) with respect to \(\displaystyle x\) and treat the other variables as constants. 

As such,

\(\displaystyle \frac{\partial }{\partial x}[F(x,y)]=\frac{\partial }{\partial x}[tan\,x-cot\,y+y^2]\)

yields

\(\displaystyle \frac{\partial F}{\partial x}=sec^2\,x\)

Example Question #1042 : Partial Derivatives

Find \(\displaystyle \frac{\partial F}{\partial x}\) given

\(\displaystyle F(x,y)=sec\,x+csc\,y\)

Possible Answers:

\(\displaystyle \frac{\partial F}{\partial x}=-csc(x)cot(x)\)

\(\displaystyle \frac{\partial F}{\partial x}=sec(x)tan(x)\)

\(\displaystyle \frac{\partial F}{\partial x}=sec^2\,x\)

Correct answer:

\(\displaystyle \frac{\partial F}{\partial x}=sec(x)tan(x)\)

Explanation:

To find \(\displaystyle \frac{\partial F}{\partial x}\) we take the derivative of the function \(\displaystyle F(x,y)\) with respect to \(\displaystyle x\) and treat the other variables as constants. 

As such,

\(\displaystyle \frac{\partial }{\partial x}[F(x,y)]=\frac{\partial }{\partial x}[sec\,x+csc\,y]\)

yields

\(\displaystyle \frac{\partial F}{\partial x}=sec(x)tan(x)\)

Example Question #1047 : Partial Derivatives

Find \(\displaystyle f_{yzyz}\) of the following function:

\(\displaystyle f(x,y,z)=z^3y^4+xz^2y\)

 

Possible Answers:

\(\displaystyle 12y^3z^2+2xz\)

\(\displaystyle 36y^2z\)

\(\displaystyle 72y^2z\)

\(\displaystyle 36y^2z^2\)

\(\displaystyle 0\)

Correct answer:

\(\displaystyle 72y^2z\)

Explanation:

To find the given partial derivative of the function, we must treat the other variable(s) as constants. For higher order partial derivatives, we work from left to right for the given variables.

First, we must find the partial derivative of the function with respect to y:

\(\displaystyle f_y=4y^3z^3+xz^2\)

Next, we find the partial derivative of this function with respect to z:

\(\displaystyle f_{yz}=12y^3z^2+2xz\)

Now, we find the partial derivative of this function with respect to y:

\(\displaystyle f_{yzy}=36y^2z^2\)

Finally, we take the partial derivative of this function with respect to z:

\(\displaystyle f_{yzyz}=72y^2z\)

Example Question #1042 : Partial Derivatives

Find \(\displaystyle f_{xy}\) of the function \(\displaystyle x^3\sin(y)+xyz\)

Possible Answers:

\(\displaystyle -3x^2\cos(y)+z\)

\(\displaystyle 4x^2\cos(y)+2z\)

\(\displaystyle 3x^2\cos(y)-z\)

\(\displaystyle 3x^2\cos(y)+z\)

Correct answer:

\(\displaystyle 3x^2\cos(y)+z\)

Explanation:

To find \(\displaystyle f_{xy}\) of the function, you must take two consecutive partial derivatives:

\(\displaystyle \frac{\partial }{\partial x}(x^3\sin(y)+xyz)=3x^2\sin(y)+yz\)

\(\displaystyle \frac{\partial }{\partial y}(3x^2\sin(y)+yz)=3x^2\cos(y)+z\)

Example Question #3411 : Calculus 3

Find \(\displaystyle f_{xy}\) of the function \(\displaystyle e^x+x\cos(y)\)

Possible Answers:

\(\displaystyle e^y-\sin(y)\)

\(\displaystyle -\sin(y)\)

\(\displaystyle ye^y-\sin(y)\)

\(\displaystyle \sin(y)\)

Correct answer:

\(\displaystyle -\sin(y)\)

Explanation:

To find \(\displaystyle f_{xy}\) of the function, you must take two consecutive partial derivatives:

\(\displaystyle \frac{\partial }{\partial x}(e^x+x\cos(y))=e^x+\cos(y)\)

\(\displaystyle \frac{\partial }{\partial y}(e^x+\cos(y))=-\sin(y)\)

Example Question #1052 : Partial Derivatives

Calculate the partial derivative with respect to \(\displaystyle x\) of the following function:

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

Possible Answers:

\(\displaystyle \frac{\partial f}{\partial x}=2xy+\frac{x}{(x+y)^2}\)

\(\displaystyle \frac{\partial f}{\partial x}=2xy^2+\frac{1}{(x+y)^2}\)

\(\displaystyle \frac{\partial f}{\partial x}=x(2xy-\frac{1}{(x+y)^2})\)

\(\displaystyle \frac{\partial f}{\partial x}=2xy-\frac{y}{(x+y)^2}\)

\(\displaystyle \frac{\partial f}{\partial x}=y(2xy+\frac{1}{(x+y)^2})\)

Correct answer:

\(\displaystyle \frac{\partial f}{\partial x}=y(2xy+\frac{1}{(x+y)^2})\)

Explanation:

When calculating the partial derivative with respect to the variable \(\displaystyle x\) of a function \(\displaystyle f\) of more than one variable, apply the standard rules for differentiating a function \(\displaystyle f(x)\) of a single variable, and treat the other variables as constants. In this case, we have:

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

We are being asked to differentiate \(\displaystyle f(x,y)\) with respect to \(\displaystyle x\), so we treat the variable \(\displaystyle y\) as a constant and apply the sum and quotient rules of differentiation to find the partial derivative \(\displaystyle \frac{\partial f}{\partial x}\), as shown:

\(\displaystyle \frac{\partial f}{\partial x}=2x(y^2)+\frac{(1)(x+y)-(1)(x)}{(x+y)^2}\)

\(\displaystyle =2xy^2+\frac{x+y-x}{(x+y)^2}\)

\(\displaystyle =2xy^2+\frac{y}{(x+y)^2}\)

Since \(\displaystyle y\) is treated as a constant, we can factor out \(\displaystyle y\) from the sum of these two terms and simplify the expression:

\(\displaystyle \frac{\partial f}{\partial x}=y(2xy+\frac{1}{(x+y)^2})\)

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