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
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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