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Solution assignment 29 Chain rule

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Assignment 29

Differentiate:

y = e^{\sin (\sqrt x )}

Solution

We need to apply the informal chain rule twice because the exponent of the exponential function is a nested function (the 'quote' means: the derivative of):

y' = e^{\sin (\sqrt x )} \cdot[\sin (\sqrt x )]'

We have:

[\sin (\sqrt x )]' = \cos (\sqrt x )\cdot[\sqrt x ]' = \cos (\sqrt x )\cdot\frac{1}{2}x^{ - \frac{1}{2}} = \displaystyle\frac{{\cos (\sqrt x )}}{{2\sqrt x }}

which results in:

y' = e^{\sin (\sqrt x )} \displaystyle\frac{{\cos (\sqrt x )}}{{2\sqrt x }}

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