# If f(x)= - e^(5x  and g(x) = 2x^3 , how do you differentiate f(g(x))  using the chain rule?

Dec 4, 2016

#### Answer:

$\left(f \left(g \left(x\right)\right)\right) ' = - 30 {x}^{2} {e}^{10 {x}^{3}}$

#### Explanation:

$f \left(g \left(x\right)\right)$ is a composite function, so differentiating it is
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determined by applying chain rule.
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Chain rule :
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$\left(f \left(g \left(x\right)\right)\right) ' = f ' \left(g \left(x\right)\right) \times g ' \left(x\right)$
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Let us compute " "color(blue)(f'(g((x)))
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Knowing the differentiation of ${e}^{u}$:
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$\left({e}^{u}\right) ' = u ' {e}^{u}$
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$f ' \left(x\right) = - 5 {e}^{5 x} \text{ }$
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then
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f'(g(x)))=-5e^(5(g(x)))
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$\textcolor{b l u e}{f ' \left(g \left(x\right)\right)} = - 5 {e}^{5 \left(2 {x}^{3}\right)}$
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$\textcolor{b l u e}{f ' \left(g \left(x\right)\right) = - 5 {e}^{10 {x}^{3}}}$
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Differentiation of $g \left(x\right)$
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$g ' \left(x\right)$ is determined by applying the power rule.
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Power Rule:
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$\textcolor{red}{\left({x}^{n}\right) ' = n {x}^{n - 1}}$
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$g ' \left(x\right) = 2 \left({x}^{3}\right) ' = 2 \left(\textcolor{red}{3 {x}^{2}}\right) = 6 {x}^{2}$
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Therefore ,
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$\left(f \left(g \left(x\right)\right)\right) ' = f ' \left(g \left(x\right)\right) \times g ' \left(x\right)$
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$\left(f \left(g \left(x\right)\right)\right) ' = - 5 {e}^{10 {x}^{3}} \times 6 {x}^{2}$
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$\left(f \left(g \left(x\right)\right)\right) ' = - 30 {x}^{2} {e}^{10 {x}^{3}}$