Key to "Given, y=tan⁡(sin−1x) then find dydx." - Plainmath

TokNeekCepTdh

Answered question

2021-11-19

Given,

y = \tan (\sin^{- 1} x)

then find

\frac{d y}{d x}

.

Answer & Explanation

Liek1993

Beginner2021-11-20Added 13 answers

Step 1
The given function is:

y = \tan (\sin^{- 1} x)

To differentiate this function, apply the chain rule of the derivatives,
Which is given by:

\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \frac{d}{d x} [g (x)]

And use,

\frac{d}{d x} (\tan x) = \sec^{2} x and \frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}

Step 2

y = \tan (\sin^{- 1} x)

\frac{d y}{d x} = \frac{d}{d x} [\tan (\sin^{- 1} x)]

\frac{d y}{d x} = \sec^{2} (\sin^{- 1} x) \cdot \frac{d}{d x} (\sin^{- 1} x)

\frac{d y}{d x} = \sec^{2} (\sin^{- 1} x) \cdot \frac{1}{\sqrt{1 - x^{2}}}

\frac{d y}{d x} = \frac{\sec^{2} (\sin^{- 1} x)}{\sqrt{1 - x^{2}}}

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