Find dy - dx if y- sin -1 - 6x-4- 1-4 x 2 - 5 -

y=sin^ 1 [ 6x 4√(1 4x^2)5 ] Multiple sin on both the sides, we get sin(y)=6x 4√(1 4x^2)5 Differentiate both sides with respect to x, cos(y)dydx=65 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

Find dy / d...

Question

Find $\frac{d y}{d x}$ if $y = {s i n}^{- 1} [\frac{6 x - 4 \sqrt{1 - 4 x^{2}}}{5}] .$

Open in App

Solution

$y = s i n^{- 1} [\frac{6 x - 4 \sqrt{1 - 4 x^{2}}}{5}]$
Multiple sin on both the sides, we get
$s i n (y) = \frac{6 x - 4 \sqrt{1 - 4 x^{2}}}{5}$
Differentiate both sides with respect to x,
$c o s (y) \frac{d y}{d x} = \frac{6}{5} - \frac{4}{5} [\frac{- 8 x}{2 \sqrt{1 - 4 x^{2}}}]$
$\sqrt{1 - s i n^{2} y} \frac{d y}{d x} = \frac{6}{5} - \frac{4}{5} [\frac{- 4 x}{\sqrt{1 - 4 x^{2}}}]$
$\sqrt{1 - (\frac{6 x - 4 \sqrt{1 - 4 x^{2}}}{5})^{2}} \frac{d y}{d x} = \frac{6}{5} - \frac{4}{5} [\frac{- 4 x}{\sqrt{1 - 4 x^{2}}}]$
$\frac{d y}{d x} = \frac{16 x + 6 \sqrt{1 - 4 x^{2}}}{\sqrt{1 - 4 x^{2}} \sqrt{25 - ((6 x - 4 \sqrt{1 - x^{2}})^{2})}} .$

Suggest Corrections

Similar questions

Q. Find

\frac{d y}{d x}

for

y = {sin}^{- 1} (\frac{6 x - 4 \sqrt{1 - 4 x^{2}}}{5})

$F i n d \frac{d y}{d x}, i f y = s i n^{- 1} x + s i n^{- 1} \sqrt{1 - x^{2}}, - 1 \leq x, \leq 1.$

If x cos(a+y) = cos y, then prove that $\frac{d y}{d x} = \frac{c o s^{2} (a + y)}{s i n a}$

Hence show that sin a $\frac{d^{2} y}{d x^{2}} + s i n 2 (a + y) \frac{d y}{d x} = 0$

OR Find $\frac{d y}{d x}$ if $y = s i n^{- 1} [\begin{matrix} \frac{6 x - 4 \sqrt{1 - 4 x^{2}}}{5} \end{matrix}] .$

Q. If

y = s i n^{- 1} (6 x \sqrt{1 - 9 x^{2}})

, then

\frac{d y}{d x}

Q. If

y = (x^{2} + 1)^{5} sin 4 x

, find

\frac{d y}{d x}

Join BYJU'S Learning Program

Find dydx if y=sin−1[6x−4√1−4x25].

Find $\frac{d y}{d x}$ if $y = {s i n}^{- 1} [\frac{6 x - 4 \sqrt{1 - 4 x^{2}}}{5}] .$