Differentiation of function with chain rule

Question

the following expression is part of a function I have to differentiate:

$y = \tan^3(5x^4-7)$

I tried using the chain-rule, so:

$ y' = 3\tan^2(5x^4-7)\cdot(20x^3)$

is this correct?

Answer 1

Not quite correct; we need to also differentiate the "tangent" component.

In general, the chain rule for a composition of functions $y = f(g(x))$ is given by: $$y'= f'(g(x)) g'(x).\,$$

In your case, we actually have $y = \Big(f(g(x)\Big)^3$, which as you correctly noticed, involves the power rule, as well. So we have $$g(x) = 5x^4 - 7 \implies g'(x) = 20x^3$$ $$\;f(g(x)) = \tan(g(x))\implies f'(g(x)) = \sec^2(g(x)) = \sec^2(5x^4 - 7)$$

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Putting it all together gives us:

$$\begin{align} y = \Big(f(g(x)\Big)^3 \implies y' & = 3\Big(f(g(x)\Big)^2\cdot f'(g(x))\cdot g'(x) \\ \\ & = 3\Big(\tan(5x^4 - 7)\Big)^2\cdot \sec^2(5x^4 - 7)\cdot (20x^3) \\ \\ & = 60x^3 \tan^2(5x^4 - 7)\sec^2(5x^4 - 7)\end{align}$$

Answer 2

Set $t=5x^4-7,~~~u=\tan(t)$ so, you have $$y=u^3$$. Now use the following routine formulas:

$$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dt}\cdot\frac{dt}{dx},~~~(\tan (t))'=1+\tan^2(t)$$