skeexerxo175o

2021-11-21

A table of values for f,g, f', and g' is given.
$\begin{array}{|ccccc|}\hline x& f\left(x\right)& g\left(x\right)& {f}^{\prime }\left(x\right)& {g}^{\prime }\left(x\right)\\ 1& 3& 2& 4& 6\\ 2& 1& 8& 5& 7\\ 3& 7& 2& 7& 9\\ \hline\end{array}$
a) If h(x)=f(g(x)), find h'(3)
b) If H(x)=g(f(x)),find H'(1).

### Answer & Explanation

Lible1953

Given:
To determine:
a) $h\left(x\right)=f\left(g\left(x\right)\right)$
To find h'(3)
$h\left(x\right)=f\left(g\left(x\right)\right)$
Then,${h}^{\prime }\left(x\right)={f}^{\prime }\left(g\left(x\right)\right)\cdot {g}^{\prime }\left(x\right)$
Putting x=3,
${h}^{\prime }\left(3\right)={f}^{\prime }\left(g\left(3\right)\right)\cdot {g}^{\prime }\left(3\right)$
From the given table,
$g\left(3\right)=2$ and ${g}^{\prime }\left(3\right)=9$
So, ${h}^{\prime }\left(3\right)={f}^{\prime }\left(2\right)\cdot 9$
From the table, ${f}^{\prime }\left(2\right)=5$
${h}^{\prime }\left(3\right)=5\cdot 9$
${h}^{\prime }\left(3\right)=45$

Fearen

b) H(x)=g(f(x))
Then,
${H}^{\prime }\left(x\right)={g}^{\prime }\left(f\left(x\right)\right)\cdot {f}^{\prime }\left(x\right)$
Putting x=1
${H}^{\prime }\left(1\right)={g}^{\prime }\left(f\left(1\right)\right)\cdot {f}^{\prime }\left(1\right)$
From the given table, $f\left(1\right)=3$ and ${f}^{\prime }\left(1\right)=4$
${H}^{\prime }\left(1\right)=9\cdot 4$
${H}^{\prime }\left(1\right)=36$

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