# Multivariable Chain Rule – Exercise 3327

Exercise

Given

$$u(x,y)=\ln (e^x+e^y)$$

$$v(x)=x^3$$

$$U(t)=u(x,v(x))$$

Calculate the derivative

$$U'(x)$$

$$U'(x)=\frac{e^x}{e^x+e^{x^3}}+\frac{3x^2e^{x^3}}{e^x+e^{x^3}}$$

Solution

We will use the chain rule to calculate the derivative

$$U'(x)=u'_x+u'_v\cdot v'_x$$

We have

$$U(x)=u(x,v(x))$$

And function u is

$$u(x,y)=\ln (e^x+e^y)$$

We put it in the function and get

$$u(x,v)=\ln (e^x+e^{v})$$

We calculate the partial derivatives of u.

$$u'_x=\frac{1}{e^x+e^v}\cdot e^x=$$

$$=\frac{e^x}{e^x+e^v}$$

$$u'_v=\frac{1}{e^x+e^v}\cdot e^v=$$

$$=\frac{e^v}{e^x+e^v}$$

We calculate the derivative of v.

$$v'_x=3x^2$$

We put the derivatives in U’ and get

$$U'(x)=u'_x+u'_v\cdot v'_x=$$

$$=\frac{e^x}{e^x+e^v}+\frac{e^v}{e^x+e^v}\cdot 3x^2$$

$$=\frac{e^x}{e^x+e^{x^3}}+\frac{e^{x^3}}{e^x+e^{x^3}}\cdot 3x^2$$

$$=\frac{e^x}{e^x+e^{x^3}}+\frac{3x^2e^{x^3}}{e^x+e^{x^3}}$$

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