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Q55E

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Essential Calculus: Early Transcendentals
Found in: Page 121
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

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 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.

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

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