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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

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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