# Calculus/Chain Rule/Solutions

1. Evaluate ${\displaystyle f'(x)}$ if ${\displaystyle f(x)=(x^{2}+5)^{2}}$, first by expanding and differentiating directly, and then by applying the chain rule on ${\displaystyle f(u(x))=u^{2}}$ where ${\displaystyle u=x^{2}+5}$. Compare answers.
First method:
${\displaystyle f(x)=x^{4}+10x^{2}+25}$
${\displaystyle \mathbf {f'(x)=4x^{3}+20x} }$

Second method:

${\displaystyle f'(u(x))={\frac {df}{du}}\cdot {\frac {du}{dx}}=2u\cdot 2x=2(x^{2}+5)\cdot 2x=\mathbf {4x^{3}+20x} }$
The two methods give the same answer.
First method:
${\displaystyle f(x)=x^{4}+10x^{2}+25}$
${\displaystyle \mathbf {f'(x)=4x^{3}+20x} }$

Second method:

${\displaystyle f'(u(x))={\frac {df}{du}}\cdot {\frac {du}{dx}}=2u\cdot 2x=2(x^{2}+5)\cdot 2x=\mathbf {4x^{3}+20x} }$
The two methods give the same answer.
2. Evaluate the derivative of ${\displaystyle y={\sqrt {1+x^{2}}}}$ using the chain rule by letting ${\displaystyle y={\sqrt {u}}}$ and ${\displaystyle u=1+x^{2}}$.
:${\displaystyle {\frac {dy}{du}}={\frac {1}{2{\sqrt {u}}}};\quad {\frac {du}{dx}}=2x}$
${\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}={\frac {1}{2{\sqrt {1+x^{2}}}}}\cdot 2x=\mathbf {\frac {x}{\sqrt {1+x^{2}}}} }$
:${\displaystyle {\frac {dy}{du}}={\frac {1}{2{\sqrt {u}}}};\quad {\frac {du}{dx}}=2x}$
${\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}={\frac {1}{2{\sqrt {1+x^{2}}}}}\cdot 2x=\mathbf {\frac {x}{\sqrt {1+x^{2}}}} }$