Question

13 mark for review
the table above gives values of the differentiable functions f and g and their derivatives at x = 1. if h(x)=(2f(x)+3)(1 + g(x)), then h(1)=
a -28
b -16
c 40
d 44
e 47

Explanation:

Step1: Recall product - rule

The product - rule states that if $h(x)=u(x)v(x)$, then $h^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$. Here, $u(x)=2f(x)+3$ and $v(x)=1 + g(x)$.
First, find $u^{\prime}(x)$ and $v^{\prime}(x)$. Since $u(x)=2f(x)+3$, then $u^{\prime}(x)=2f^{\prime}(x)$ by the constant - multiple and sum rules of differentiation. And since $v(x)=1 + g(x)$, then $v^{\prime}(x)=g^{\prime}(x)$.

Step2: Evaluate $u(1)$, $u^{\prime}(1)$, $v(1)$ and $v^{\prime}(1)$

Given $f(1) = 3$, then $u(1)=2f(1)+3=2\times3 + 3=6 + 3=9$.
Given $f^{\prime}(1)=-2$, then $u^{\prime}(1)=2f^{\prime}(1)=2\times(-2)=-4$.
Given $g(1)=-3$, then $v(1)=1 + g(1)=1+( - 3)=-2$.
Given $g^{\prime}(1)=4$, then $v^{\prime}(1)=g^{\prime}(1)=4$.

Step3: Apply the product - rule to find $h^{\prime}(1)$

$h^{\prime}(1)=u^{\prime}(1)v(1)+u(1)v^{\prime}(1)$.
Substitute the values: $h^{\prime}(1)=(-4)\times(-2)+9\times4$.
First, calculate $(-4)\times(-2)=8$ and $9\times4 = 36$.
Then $h^{\prime}(1)=8 + 36=44$.

Answer:

D. 44