If fx=sin-13x-4x3, then the value of f'(0) is

Solution:

The given function is $f (x) = \sin^{- 1} (3 x - 4 x^{3})$ .

It is known that $\sin 3 θ = 3 \sin θ - 4 \sin^{3} θ$ .

Put $x = \sin θ$ in the given function.

$f (x) = \sin^{- 1} (3 x - 4 x^{3}) f (x) = \sin^{- 1} (3 \sin θ - 4 {(\sin θ)}^{3}) f (x) = \sin^{- 1} (3 \sin θ - 4 \sin^{3} θ) f (x) = \sin^{- 1} (\sin 3 θ) f (x) = 3 θ f (x) = 3 \sin^{- 1} x$

Differentiating the function,

$f' (x) = 3 \times \frac{1}{\sqrt{1 - x^{2}}} f' (x) = \frac{3}{\sqrt{1 - x^{2}}}$

At $x = 0$ ,

$f' (0) = \frac{3}{\sqrt{1 - 0}} f' (0) = 3$

Hence, the value of $f' (0)$ is $3$ .

If $f (x) = \sin^{- 1} (3 x - 4 x^{3})$ , then the value of $f^{'} (0)$ is

Solution:

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