Question

Find the derivative of tan⁻¹ x/√1-x² w.r.t. sin⁻¹ (2x√1-x²). 0 < x < 1/√2

Answer 1

Answer:

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Answer 2

Answer:

$\frac{1}{2}$

Step-by-step explanation:

We have to find the value of $\frac{d(tan^{-1}(\frac{x}{\sqrt{1-x^{2}}}))}{d(sin^{-1}(2x\sqrt{1-x^{2}}))}$ =D (say)

Now, let us assume that, $x=sin\alpha$.

Then, $tan^{-1}[\frac{x}{\sqrt{1-x^{2} } } ] =tan^{-1}[\frac{sin\alpha }{cos\alpha } ] =tan^{-1}[tan\alpha ]=\alpha$....... (1)

Again, $sin^{-1}[2x\sqrt{1-x^{2} } ] = sin^{-1}[2sin\alpha.\sqrt{1-sin^{2}\alpha  } ]=sin^{-1}[sin2\alpha ]= 2\alpha$............(2)

Hence, we can write, D=$\frac{d(tan^{-1}(\frac{x}{\sqrt{1-x^{2}}}))}{d(sin^{-1}(2x\sqrt{1-x^{2}}))}$

=$\frac{d\alpha }{d(2\alpha )}$ {From equation (1) and (2)}

= $\frac{1}{\frac{d(2\alpha )}{d\alpha } }$

=$\frac{1}{2}$ (Answer)