How do you take the derivative of ${tan}^{10} 5 x$ $tan^10 5x$ ?

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Answer 1 · 2015-06-17T00:17:10

Consider the regular derivative of $tan u$ $tanu$ .

$\frac{d}{d x} [tan u] = {sec}^{2} u \cdot (\frac{d u}{d x})$ $d/(dx)[tanu] = sec^2u*((du)/(dx))$

Since $u (x) = 5 x$ $u(x) = 5x$ and we have a power function:

$\frac{d}{d x} [{(tan u)}^{n}] = n {(tan u)}^{n - 1} \cdot {sec}^{2} u \cdot (\frac{d u}{d x})$ $d/(dx)[(tanu)^n] = n(tanu)^(n-1)*sec^2u*((du)/(dx))$

$\frac{d}{d x} [{(tan (5 x))}^{10}] = 10 {(tan (5 x))}^{9} \cdot {sec}^{2} (5 x) \cdot 5$ $d/(dx)[(tan(5x))^(10)] = 10(tan(5x))^9*sec^2(5x)*5$

$= 50 {tan}^{9} (5 x) {sec}^{2} (5 x)$ $= 50tan^9(5x)sec^2(5x)$

How do you take the derivative of tan105xtan^10 5x?