Differentials - AP Calculus BC

Card 0 of 145

Question

Find the total differential, , of the following function

Answer

The total differential is defined as

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

For the function

We first find

$\frac{\partial z}{\partial x}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial x}=1-sin,x$

We then find

$\frac{\partial z}{\partial y}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial y}=cos,y+1$

We then substitute these partial derivatives into the first equation to get the total differential

Compare your answer with the correct one above

Question

Compute the differentials for the following function.

$z=e^{x^2+4y^2}\sin(y)$

Answer

What we need to do is take derivatives, and remember the general equation.

$z=e^{x^2+4y^2}\sin(y)$

When taking the derivative with respect to y recall that the product rule needs to be used.

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

Compare your answer with the correct one above

Question

Find the total differential, , of the function

$z=e^{x^2+y^2}+sec(2x)$

Answer

The total differential is defined as

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

We first find $\frac{\partial z}{\partial x}$ by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}[e^{x^2+y^2}+sec(2x)]=2xe^{x^2+y^2}+2sec(2x)tan(2x)$

We then find $\frac{\partial z}{\partial y}$ by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}[e^{x^2+y^2}+sec(2x)]=2ye^{x^2+y^2}$

We then substitute these partial derivatives into the first equation to get the total differential

$dz=[2xe^{x^2+y^2}+2sec(2x)tan(2x)]dx+2ye^{x^2+y^2}dy$

Compare your answer with the correct one above

Question

Find the total differential, , of the function

$w=e^{x+y}tan(4z)+sin(z)$

Answer

The total differential is defined as

$dw=\frac{\partial w}{\partial x}dx+\frac{\partial w}{\partial y}dy+\frac{\partial w}{\partial z}dz$

We first find $\frac{\partial w}{\partial x}$ by taking the derivative with respect to and treating the other variables as constants.

$\frac{\partial w}{\partial x}=\frac{\partial }{\partial x}[e^{x+y}tan(4z)+sin(z)]=e^{x+y}tan(4z)$

We then find $\frac{\partial w}{\partial y}$ by taking the derivative with respect to and treating the other variables as constants.

$\frac{\partial w}{\partial y}=\frac{\partial }{\partial y}[e^{x+y}tan(4z)+sin(z)]=e^{x+y}tan(4z)$

We then find $\frac{\partial w}{\partial z}$ by taking the derivative with respect to and treating the other variables as constants.

$\frac{\partial w}{\partial z}=\frac{\partial }{\partial y}[e^{x+y}tan(4z)+sin(z)]=4e^{x+y}sec^2(4z)+cos(z)$

We then substitute these partial derivatives into the first equation to get the total differential

$dw=e^{x+y}tan(4z)dx+e^{x+y}tan(4z)}dy +[4e^{x+y}sec^2(4z)+cos(z)]dz$

Compare your answer with the correct one above

Question

Find the total differential , , of the function

Answer

The total differential is defined as

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

We first find

$\frac{\partial z}{\partial x}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial x}=e^x$

We then find

$\frac{\partial z}{\partial y}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial y}=-sin(y)$

We then substitute these partial derivatives into the first equation to get the total differential

Compare your answer with the correct one above

Question

Find the total derivative of the function:

$f(x, y)=x\sec(y)+6y^4$

Answer

The total derivative of a function of two variables is given by the following:

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives of the function are

$f_x=\sec(y)$

$f_y=x\sec(y)\tan(y)+24y^3$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

So, our final answer is

$\sec(y)dx+(x\sec(y)\tan(y)+24y^3)dy$

Compare your answer with the correct one above

Question

Find the total derivative of the function:

$f(x, y)=\sqrt{x+y^2}+\ln(x^2)$

Answer

The total derivative of a function of two variables is given by

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

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

$f_y=\frac{y}{\sqrt{x+y^2}}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$

Compare your answer with the correct one above

Question

Find the total derivative of the function:

$f(x, y,z)=\cos(xy^2)+e^{z^2}$

Answer

The total derivative of a function is given by

So, we must find the partial derivatives. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partials are

$f_x=-y^2\sin(xy^2)$

$f_y=-2xy\sin(xy^2)$

$f_z=2ze^{z^2}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$

Compare your answer with the correct one above

Question

Find the total derivative of the following function:

$f(x, y,z)=x^2y+\csc(z)$

Answer

The total derivative of a function is given by

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_z=-\csc(z)\cot(z)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

Compare your answer with the correct one above

Question

Find the differential of the following function:

$f(x, y, z)=x^5+\tan(y^2z)$

Answer

The differential of a function is given by

So, we must find the partial derivatives of the function. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_y=2yz\sec^2(y^2z)$

$f_z=y^2\sec^2(y^2z)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Compare your answer with the correct one above

Question

Find the differential of the function

$f(x, y, z)=ze^y+\ln(x^2)$

Answer

The differential of a function is given by

The partial derivatives of the function are

$f_x=\frac{2}{x}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Compare your answer with the correct one above

Question

Find the differential of the function:

Answer

The differential of the function is given by

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

Compare your answer with the correct one above

Question

Find the total differential, , of the following function

Answer

The total differential is defined as

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

For the function

We first find

$\frac{\partial z}{\partial x}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial x}=y+2$

We then find

$\frac{\partial z}{\partial y}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial y}=x+2$

We then substitute these partial derivatives into the first equation to get the total differential

Compare your answer with the correct one above

Question

Find the total differential, , of the following function

Answer

The total differential is defined as

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

For the function

We first find

$\frac{\partial z}{\partial x}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial x}=2xy^2+e^x$

We then find

$\frac{\partial z}{\partial y}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial y}=2x^2y-sin,y$

We then substitute these partial derivatives into the first equation to get the total differential

Compare your answer with the correct one above

Question

Find the differential of the function:

$k=z^3\tan(y^2)$

Answer

The differential of a function is given by

The partial derivatives of the function are

$f_y=2yz^3\sec^2(y^2)$

$f_z=3z^2\tan(y^2)$

Compare your answer with the correct one above

Question

Find the differential of the function:

Answer

The differential of the function is given by

The partial derivatives are

Compare your answer with the correct one above

Question

If , calculate the differential when moving from the point to the point .

Answer

The equation for is

.

Evaluating partial derivatives and substituting, we get

$dx = \Delta x = 1, dy=\Delta y = 2$

Plugging in, we get

.

Compare your answer with the correct one above

Question

If $f(x,y) = \ln(xy)$ , calculate the differential when moving from the point to the point .

Answer

The equation for is

.

Evaluating partial derivatives and substituting, we get

$f_x(x_0,y_0) = f_x(1,1) = \frac{1}{(1)} =1$

$f_y(x_0,y_0) = f_y(1,1) = \frac{1}{(1)}=1$

$dx = \Delta x = e-1, dy=\Delta y = e-1$

Plugging in, we get

Compare your answer with the correct one above

Question

If , calculate the differential when moving from to.

Answer

The equation for is

.

Evaluating partial derivatives and substituting, we get

$f_x(x_0,y_0) = f_x(1,0) = e^{(0)} =1$

$f_y(x_0,y_0) = f_y(1,0) = (1)e^{(0)}=1$

$dx = \Delta x = -1, dy=\Delta y = 1$

Plugging in, we get

.

Compare your answer with the correct one above

Question

Find the differential of the following function:

Answer

The differential of the function is given by

The partial derivatives are

, ,

Compare your answer with the correct one above

Tap the card to reveal the answer