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Q55E

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Found in: Page 121

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# A table of values for$$f,g,{f^\prime }$$, and $${g^\prime }$$is given.(a) If$$h(x) = f(g(x))$$, find$${h^\prime }(1)$$.(b) If$$H(x) = g(f(x))$$, find$${H^\prime }(1)$$.

(a) If $$h(x) = f(g(x))$$, $${h^\prime }(1) = 30$$.

(b) If $$H(x) = g(f(x))$$, $${H^\prime }(1) = 36$$.

See the step by step solution

## Step 1: Given Information.

The table of values for $$x = 1,2,3$$.

## Step 2: Definition of chain rule.

Chain rule is a rule which defines a way to find the derivative of a given function.

$$\frac{{d(f(g(x)))}}{{dx}} = \frac{{d(f(g(x)))}}{{d(g(x))}} \cdot \frac{{d(g(x))}}{{dx}}$$

## Step 3:Use chain rule for differentiation for $$h(x) = f(g(x))$$.

State the chain rule.

$$\frac{{d(f(g(x)))}}{{dx}} = \frac{{d(f(g(x)))}}{{d(g(x))}} \cdot \frac{{d(g(x))}}{{dx}}$$

Here,

$${h^\prime }(x) = {f^\prime }(g(x)) \cdot {g^\prime }(x)$$

Substitute $$x = 1$$ and then substitute the tabular values to the obtained answer.

$$\begin{array}{c}{h^\prime }(1) &=& {f^\prime }(g(1)) \cdot {g^\prime }(1)\\{h^\prime }(1) &=& {f^\prime }(2) \cdot 6\\{h^\prime }(1) &=& 5 \cdot 6\\ &=& 30\end{array}$$

## Step 4: Use chain rule for differentiation for $$H(x) = g(h(x))$$.

State the chain rule.

$$\frac{{d(f(g(x)))}}{{dx}} = \frac{{d(f(g(x)))}}{{d(g(x))}} \cdot \frac{{d(g(x))}}{{dx}}$$

Here,

$${H^\prime }(x) = {g^\prime }(f(x)) \cdot {f^\prime }(x)$$

Substitute $$x = 1$$ and then substitute the tabular values to the obtained answer.

$$\begin{array}{c}{H^\prime }(1) &=& {g^\prime }(f(1)) \cdot {f^\prime }(1)\\{H^\prime }(1) &=& {g^\prime }(3) \cdot 4\\{H^\prime }(1) &=& 9 \cdot 4\\ &=& 36\end{array}$$