Question

suppose both $f$ and $g$ are differentiable with $g(2)=3, g(2)=5$, and $f(6)=4$. consider the function $h(x)=f(xg(x))$. then $h(2)=$

question 5
2 pts
suppose $g(x)$ is differentiable with $g(1)=2$ and $g(1)= - 3$. find an equation for the tangent line to the graph of the function $f(x)=(g(x^{2})+x)^{3}$ at the point where $x = 1$. the answer is $y=mx + b$ where $m=$
$b=$

Explanation:

Step1: Apply chain - rule to find $h'(x)$

By the chain - rule, if $h(x)=f(xg(x))$, then $h'(x)=f'(xg(x))\cdot(xg'(x)+g(x))$.

Step2: Evaluate $h'(2)$

Substitute $x = 2$ into $h'(x)$. We know that $g(2)=3$, $g'(2)=5$, and $f'(6)=4$. When $x = 2$, $xg(x)=2\times g(2)=2\times3 = 6$. Then $h'(2)=f'(2g(2))\cdot(2g'(2)+g(2))=f'(6)\cdot(2\times5 + 3)$.

Step3: Calculate the value of $h'(2)$

Since $f'(6)=4$, we have $h'(2)=4\times(10 + 3)=4\times13=52$.

Step4: For the second part, find $f'(x)$ using chain - rule and sum - rule

Let $u(x)=g(x^{2})+x$. Then $f(x)=u(x)^{3}$. By the chain - rule, $f'(x)=3u(x)^{2}\cdot u'(x)$. And $u'(x)=g'(x^{2})\cdot2x + 1$.

Step5: Evaluate $u(1)$ and $u'(1)$

When $x = 1$, $u(1)=g(1^{2})+1=g(1)+1$. Since $g(1)=2$, $u(1)=2 + 1=3$.
$u'(1)=g'(1^{2})\cdot2\times1+1=g'(1)\cdot2 + 1$. Since $g'(1)=-3$, $u'(1)=-3\times2 + 1=-6 + 1=-5$.

Step6: Calculate $f'(1)$

$f'(1)=3u(1)^{2}\cdot u'(1)=3\times3^{2}\times(-5)=3\times9\times(-5)=-135$. Also, $f(1)=(g(1)+1)^{3}=(2 + 1)^{3}=27$.
The equation of the tangent line is $y - y_{0}=m(x - x_{0})$. Here $x_{0}=1$, $y_{0}=27$, and $m=-135$. So $y-27=-135(x - 1)$, which simplifies to $y=-135x+162$. So $m=-135$ and $b = 162$.

Answer:

$h'(2)=52$
$m=-135$
$b = 162$