Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (e)/Solution 1

Let $\displaystyle f_{1}(x)=x^{5}$ and let $\displaystyle f_{2}(x)=\sin(\cos(3x^{2}))$. Then

$\displaystyle y=f_{1}(f_{2}(x))=\sin ^{5}(\cos(3x^{2}))$.

So the derivative by the chain rule is

$\displaystyle y'=f'_{1}(f_{2}(x))f_{2}'(x)$ (Equation 1)

Now, $\displaystyle f'_{1}(x)=5x^{4}$ and so

$\displaystyle f'_{1}(f_{2}(x))=5(\sin(\cos(3x^{2})))^{4}=5\sin ^{4}(\cos(3x^{2}))$ (Equation 2)

For $\displaystyle f'_{2}(x)$, we again use the chain rule. Let $\displaystyle f_{3}(x)=\sin(x)$ and $\displaystyle f_{4}(x)=\cos(3x^{2})$ so we have that $\displaystyle f_{2}(x)=f_{3}(f_{4}(x))$. Using the chain rule, we have that

$\displaystyle f'_{2}(x)=f'_{3}(f_{4}(x))f_{4}'(x)$ (Equation 3)

Notice that $\displaystyle f_{3}'(x)=\cos(x)$ and thus

$\displaystyle f'_{3}(f_{4}(x))=\cos(\cos(3x^{2}))$ (Equation 4)

Thus it suffices to find $\displaystyle f'_{4}(x)$. Guess what we're going to use... the chain rule! Let $\displaystyle f_{5}(x)=\cos(x)$ and $\displaystyle f_{6}(x)=3x^{2}$. Then, the chain rule states that

$\displaystyle f'_{4}(x)=f'_{5}(f_{6}(x))f'_{6}(x)$ (Equation 5)

Now, $\displaystyle f'_{5}(x)=-\sin(x)$ and so

$\displaystyle f'_{5}(f_{6}(x))=-\sin(3x^{2})$ (Equation 6)

Here, the derivative of $\displaystyle f_{6}(x)$ is actually doable! We have that

$\displaystyle f'_{6}(x)={\frac {d}{dx}}(3x^{2})=6x$

and hence by Equations 5 and 6, we have

$\displaystyle f'_{4}(x)=f'_{5}(f_{6}(x))f'_{6}(x)=(-\sin(3x^{2}))(6x)$ (Equation 7)

Combining Equations 1,2,3,4,6 and 7, we have

${\begin{aligned}y'&=f'_{1}(f_{2}(x))f_{2}'(x)&&{\text{(via Equation 1)}}\\&=5\sin ^{4}(\cos(3x^{2}))(f'_{3}(f_{4}(x))f_{4}'(x))&&{\text{(via Equation 2 and 3)}}\\&=5\sin ^{4}(\cos(3x^{2}))(\cos(\cos(3x^{2}))(-\sin(3x^{2}))(6x))&&{\text{(via Equation 4 and 7)}}\\&=-30x\sin ^{4}(\cos(3x^{2}))\cos(\cos(3x^{2}))\sin(3x^{2})\end{aligned}}$

Phew!