Differentiation Rules

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Multivariable Calculus › Differentiation Rules

Questions 1 - 3

1

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

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

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

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

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

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

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

Explanation

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

Natural Log:

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

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

Exponential Functions:

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

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

Power Functions:

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

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

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

2

Find $\frac{\partial f}{dz}$ .

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

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

$\frac{\partial f}{dz}=10xyz^9+10e^z$

$\frac{\partial f}{dz}=10xyz^9$

$\frac{\partial f}{dz}=\frac{xy}{z\ln^2(yz)}$

$\frac{\partial f}{dz}=10e^z$

Explanation

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

Natural Log:

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

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

Exponential Functions:

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

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

Power Functions:

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

$\frac{\partial f}{dz}=10xyz^9-xy(\ln(yz))^{-2}(\frac{y}{yz})+10e^z+0$

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

3

Find $\frac{\partial f}{dz}$ .

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

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

$\frac{\partial f}{dz}=10xyz^9+10e^z$

$\frac{\partial f}{dz}=10xyz^9$

$\frac{\partial f}{dz}=\frac{xy}{z\ln^2(yz)}$

$\frac{\partial f}{dz}=10e^z$

Explanation

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

Natural Log:

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

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

Exponential Functions:

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

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

Power Functions:

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

$\frac{\partial f}{dz}=10xyz^9-xy(\ln(yz))^{-2}(\frac{y}{yz})+10e^z+0$

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

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