# If f(x)= tan5 x  and g(x) = -x^2 -1 , how do you differentiate f(g(x))  using the chain rule?

Jul 9, 2016

${f}^{'} \left(g \left(x\right)\right) = - 10 x {\sec}^{2} \left(5 x\right)$

#### Explanation:

Known:$\frac{d}{\mathrm{dx}} \left(\tan x\right) = {\sec}^{2} x$ so $\frac{d}{\mathrm{dx}} \left(\tan 5 x\right) = 5 {\sec}^{2} x$
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Let $g \left(x\right) = u = - {x}^{2} - 1 \to \frac{\mathrm{du}}{\mathrm{dx}} = - 2 x$

Thus $f \left(g \left(x\right)\right) = y = \tan \left(5 u\right) \to \frac{\mathrm{dy}}{\mathrm{du}} = 5 {\sec}^{2} \left(u\right)$
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However:$\text{ "(du)/(dx)xx(dy)/(du)" " =" " (du)/(du)xx(dy)/(dx)" "=" } \frac{\mathrm{dy}}{\mathrm{dx}}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x \times 5 {\sec}^{2} \left(5 x\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 10 x {\sec}^{2} \left(5 x\right)$

Or if you prefer

${f}^{'} \left(g \left(x\right)\right) = - 10 x {\sec}^{2} \left(5 x\right)$