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

$U'(x)$

Final Answer

Show final answer

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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