If f(x) =sec^2(x/2)  and g(x) = sqrt(5x-1 , what is f'(g(x)) ?

Oct 22, 2017

$f ' \left(g \left(x\right)\right) = \tan \left(\frac{\sqrt{5 x - 1}}{2}\right) \times {\sec}^{2} \left(\frac{\sqrt{5 x - 1}}{2}\right)$

Explanation:

f'(x)=??
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$f \left(x\right) \text{ }$ is a composite function composed of two functions
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Trigonometric function $u \left(x\right) = \sec \left(\frac{x}{2}\right)$ and
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Polynomial function $v \left(x\right) = {x}^{2}$
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$f ' \left(x\right)$ is determined by applying the chain rule .
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$f \left(x\right) = v \left(u \left(x\right)\right)$
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$f ' \left(x\right) = v ' \left(u \left(x\right)\right) \times u ' \left(x\right)$
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$v ' \left(x\right) = \left({x}^{2}\right) ' = 2 x$
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$u ' \left(x\right) = \left(\sec \left(\frac{x}{2}\right)\right) ' = \left(\frac{x}{2}\right) ' \times \sec ' \left(\frac{x}{2}\right)$
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$u ' \left(x\right) = \frac{1}{2} \times \tan \left(\frac{x}{2}\right) \sec \left(\frac{x}{2}\right)$
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So,$f ' \left(x\right)$ is determined by substituting the differentiated functions
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$f ' \left(x\right) = v ' \left(u \left(x\right)\right) \times u ' \left(x\right) = 2 \left(u \left(x\right)\right) \times \frac{1}{2} \times \tan \left(\frac{x}{2}\right) \sec \left(\frac{x}{2}\right)$
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$f ' \left(x\right) = 2 \sec \left(\frac{x}{2}\right) \times \frac{1}{2} \times \tan \left(\frac{x}{2}\right) \sec \left(\frac{x}{2}\right)$
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$f ' \left(x\right) = \tan \left(\frac{x}{2}\right) \times {\sec}^{2} \left(\frac{x}{2}\right)$
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$f ' \left(g \left(x\right)\right) = \tan \left(\frac{g \left(x\right)}{2}\right) \times {\sec}^{2} \left(\frac{g \left(x\right)}{2}\right)$
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$f ' \left(g \left(x\right)\right) = \tan \left(\frac{\sqrt{5 x - 1}}{2}\right) \times {\sec}^{2} \left(\frac{\sqrt{5 x - 1}}{2}\right)$