Question

X f(x) f ′(x) g(x) g′(x)

2 5 1 2 1
4 −2 −6 5 4
5 1 2 4 5
6 5 1 2 −2


The functions f and g have continuous derivatives. The table gives values of f, f ′, g, and g′ at selected values of x.

Part A: Find h′(2) if h(x) = g(f(x)). (5 points)

Part B: Find m′(2) if m(x) = f(x2). (5 points)

Part C: Let k(x) = f(g(x)). Write an equation for the line tangent to the graph of k at x = 4. (10 points)

Part D: Let j of x is equal to g of x divided by f of x. Find j′(2). (10 points)

Answer 1

A. h'(2) = 5

B. m'(2) = -24

C. y = 8x - 31

D. j'(2) = 3/25 = 0.12

Explanation:

When differentiating composite functions, use the chain rule.

`\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\$\left[f\left(g(x)\right)\right]'=f'\left(g(x)\right) \cdot g'(x)$\\\end{minipage}}`

Chain Rule: The derivative of a composite function is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function.

Part A

Using the chain rule, if h(x) = g(f(x)) then h'(x) = g'(f(x)) ⋅ f'(x).

To find h'(2), substitute x = 2 into the differentiated equation:

`\begin{aligned}h'(x)& = g'(f(x)) \cdot f'(x)\\\\\implies h'(2)& = g'(f(2)) \cdot f'(2)\\& = g'(5) \cdot 1\\& = 5 \cdot 1\\&=5\end{aligned}`

Therefore, h'(2) = 5.

Part B

Using the chain rule, if m(x) = f(x²) then m'(x) = f'(x²) ⋅ 2x.

To find m'(2), substitute x = 2 into the differentiated equation:

`\begin{aligned}m'(x) &= f'(x^2) \cdot 2x\\\\\implies m'(2) &= f'(2^2) \cdot 2(2)\\&=f'(4) \cdot 4\\&=-6 \cdot 4\\&=-24\end{aligned}`

Therefore, m'(2) = -24.

Part C

To find the slope of the tangent line to the graph of k at x = 4, substitute x = 4 into the derivative of k(x).

If k(x) = f(g(x)) then k'(x) = f'(g(x)) ⋅ g'(x).

Therefore, the slope of the tangent line is:

`\begin{aligned}k'(x) &= f'(g(x)) \cdot g'(x)\\\\\implies k'(4) &= f'(g(4)) \cdot g'(4)\\&= f'(5) \cdot 4\\&= 2 \cdot 4\\&= 8\end{aligned}`

Now calculate k(x) when x = 4:

`\begin{aligned}k(x)&=f(g(x))\\\\\implies k(4) &= f(g(4))\\&=f(5)\\&=1\end{aligned}`

To write the equation of the tangent line, substitute the found slope m = 8 and point (4, 1) into the point-slope equation:

`\begin{aligned}y-y_1&=m(x-x_1)\\y-1&=8(x-4)\\y-1&=8x-32\\y&=8x-31\end{aligned}`

Therefore, an equation for the line tangent to the graph of k at x = 4 is:

• y = 8x - 31

Part D

To find the derivative of j(x) use the quotient rule.

`\textsf{If\;\;$j(x)=(g(x))/(f(x))$\;\;then:}\\\\\\j'(x)=(f(x)g'(x)-g(x)f'(x))/((f(x))^2)`

To calculate j'(2), substitute x = 2 into the equation:

`\begin{aligned} \implies j'(2)&=(f(2)g'(2)-g(2)f'(2))/((f(2))^2)\\\\&=(5 \cdot 1-2 \cdot 1)/((5)^2)\\\\&=(5-2)/(25)\\\\&=(3)/(25)\\\\&=0.12\end{aligned}`

Therefore, j'(2) = 3/25 = 0.12.