Question

find the derivative of y with respect to x.
y = 9x^4 arcsin(9x^4)+\sqrt{1 - 81x^8}
\frac{dy}{dx}=\square

Explanation:

Step1: Apply product - rule on first term

The product - rule states that if \(u = 9x^{4}\) and \(v=\arcsin(9x^{4})\), then \((uv)^\prime=u^\prime v + uv^\prime\). First, find \(u^\prime\): \(u = 9x^{4}\), so \(u^\prime=36x^{3}\). For \(v = \arcsin(9x^{4})\), by the chain - rule, if \(t = 9x^{4}\), then \(v=\arcsin(t)\) and \(v^\prime=\frac{1}{\sqrt{1 - t^{2}}}\cdot t^\prime=\frac{36x^{3}}{\sqrt{1 - 81x^{8}}}\). So, \((9x^{4}\arcsin(9x^{4}))^\prime=36x^{3}\arcsin(9x^{4})+\frac{9x^{4}\cdot36x^{3}}{\sqrt{1 - 81x^{8}}}\).

Step2: Apply chain - rule on second term

Let \(u = 1-81x^{8}\), then \(y_2=\sqrt{u}=u^{\frac{1}{2}}\). By the chain - rule, \(y_2^\prime=\frac{1}{2\sqrt{u}}\cdot u^\prime\). Since \(u = 1 - 81x^{8}\), \(u^\prime=-648x^{7}\). So, \(y_2^\prime=\frac{-648x^{7}}{2\sqrt{1 - 81x^{8}}}=\frac{-324x^{7}}{\sqrt{1 - 81x^{8}}}\).

Step3: Find the derivative of \(y\)

\(y^\prime=(9x^{4}\arcsin(9x^{4}))^\prime+(\sqrt{1 - 81x^{8}})^\prime\).

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Answer:

\(36x^{3}\arcsin(9x^{4})\)