Question

If y=sin^-1 (10x÷1+25x^2) then find dy/dx. ​

Answer 1

$\tt \red{\frac{d}{dx} \: {y=sin^{-1 }(10x÷1+25x^2)}}$

Can be written as:-

$\tt \red{\frac{d}{dx} \left(y = \operatorname{asin}{\left(25 x^{2} + 10 x \right)}\right)}$

SOLUTION

Differentiate separately both sides of the equation (treat y as a function of x) $\tt\red {\frac{d}{dx} \left(y{\left(x \right)}\right) = \frac{d}{dx} \left(\operatorname{asin}{\left(25 x^{2} + 10 x \right)}\right)}$

Differentiate the RHS of the equation.

$\tt \red{The \: function\operatorname{asin}{\left(25 x^{2} + 10 x \right)}} \\ \tt \red{ is \: the \: composition \: f{\left(g{\left(x \right)} \right)} \: of \: } \\ \tiny \tt \red{ two \: functions \: f{\left(u \right)} = \operatorname{asin}{\left(u \right)} \: and \: g{\left(x \right)} = 25 x^{2} + 10 x}$

Apply the chain rule

$\tt \red{\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)}$

$\tiny \tt \red{\color{red}{\left(\frac{d}{dx} \left(\operatorname{asin}{\left(25 x^{2} + 10 x \right)}\right)\right)} = \color{red}{\left(\frac{d}{du} \left(\operatorname{asin}{\left(u \right)}\right) \frac{d}{dx} \left(25 x^{2} + 10 x\right)\right)}}$

The derivative of the inverse sine is

$\tiny \tt \red{\frac{d}{du} \left(\operatorname{asin}{\left(u \right)}\right) = \frac{1}{\sqrt{1 - u^{2}}}\color{red}{\left(\frac{d}{du} \left(\operatorname{asin}{\left(u \right)}\right)\right)} \frac{d}{dx} \left(25 x^{2} + 10 x\right)}$

$\tt \red{ = \color{red}{\left(\frac{1}{\sqrt{1 - u^{2}}}\right)} \frac{d}{dx} \left(25 x^{2} + 10 x\right)}$

Return to the old variable:

$\tt \red{\frac{\frac{d}{dx} \left(25 x^{2} + 10 x\right)}{\sqrt{1 - \color{red}{\left(u\right)}^{2}}} = \frac{\frac{d}{dx} \left(25 x^{2} + 10 x\right)}{\sqrt{1 - \color{red}{\left(25 x^{2} + 10 x\right)}^{2}}}}$

The derivative of a sum/difference is the sum/difference of derivatives:

$\small \tt \red{\frac{\color{red}{\left(\frac{d}{dx} \left(25 x^{2} + 10 x\right)\right)}}{\sqrt{1 - \left(25 x^{2} + 10 x\right)^{2}}} = \frac{\color{red}{\left(\frac{d}{dx} \left(25 x^{2}\right) + \frac{d}{dx} \left(10 x\right)\right)}}{\sqrt{1 - \left(25 x^{2} + 10 x\right)^{2}}}}$

Apply the constant multiple rule

$\tt \tiny \red{\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right) with \: c = 10 \: and \: f{\left(x \right)} = x}$

$\tiny \tt \red{\frac{\color{red}{\left(\frac{d}{dx} \left(10 x\right)\right)} + \frac{d}{dx} \left(25 x^{2}\right)}{\sqrt{1 - \left(25 x^{2} + 10 x\right)^{2}}} = \frac{\color{red}{\left(10 \frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(25 x^{2}\right)}{\sqrt{1 - \left(25 x^{2} + 10 x\right)^{2}}}}$

Apply the power rule

$\tiny \tt\red{\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1} \: with \: n = 1, in \: other \: words, \frac{d}{dx} \left(x\right) = 1}$

$\tiny \tt \red{\frac{10 \color{red}{\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(25 x^{2}\right)}{\sqrt{1 - \left(25 x^{2} + 10 x\right)^{2}}} = \frac{10 \color{red}{\left(1\right)} + \frac{d}{dx} \left(25 x^{2}\right)}{\sqrt{1 - \left(25 x^{2} + 10 x\right)^{2}}}}$

Apply the constant multiple rule

$\tiny \tt\red{\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \: \frac{d}{dx} \left(f{\left(x \right)}\right) with \: c = 25 \: and \: f{\left(x \right)} = x^{2}}$

$\tiny \tt \red{\frac{\color{red}{\left(\frac{d}{dx} \left(25 x^{2}\right)\right)} + 10}{\sqrt{1 - \left(25 x^{2} + 10 x\right)^{2}}} = \frac{\color{red}{\left(25 \frac{d}{dx} \left(x^{2}\right)\right)} + 10}{\sqrt{1 - \left(25 x^{2} + 10 x\right)^{2}}}}$

Apply the power rule

$\tt \red{\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1} \: with \: n = 2}$

$\tiny \tt \red{\frac{25 \color{red}{\left(\frac{d}{dx} \left(x^{2}\right)\right)} + 10}{\sqrt{1 - \left(25 x^{2} + 10 x\right)^{2}}} = \frac{25 \color{red}{\left(2 x\right)} + 10}{\sqrt{1 - \left(25 x^{2} + 10 x\right)^{2}}}}$

Simplify:

$\tiny \tt \red{\frac{50 x + 10}{\sqrt{1 - \left(25 x^{2} + 10 x\right)^{2}}} = \frac{10 \left(5 x + 1\right)}{\sqrt{- 25 x^{2} \left(5 x + 2\right)^{2} + 1}}}$

$\tiny \tt \red{Thus, \frac{d}{dx} \left(\operatorname{asin}{\left(25 x^{2} + 10 x \right)}\right) = \frac{10 \left(5 x + 1\right)}{\sqrt{- 25 x^{2} \left(5 x + 2\right)^{2} + 1}}}$

Therefore,

$\tt \red{\frac{dy}{dx} = \frac{10 \left(5 x + 1\right)}{\sqrt{- 25 x^{2} \left(5 x + 2\right)^{2} + 1}}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg(\dfrac{10x}{1 + {25x}^{2} } \bigg)$

can be rewritten as

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg(\dfrac{2 \times 5x}{1 + {(5x)}^{2} } \bigg)$

Let assume that

$\purple{\rm :\longmapsto\:5x = tanz \: \: \rm\implies \:z = {tan}^{ - 1}5x}$

So, above expression can be rewritten as

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg(\dfrac{2tanz}{1 + {tan}^{2}z} \bigg)$

$\rm :\longmapsto\:y = {sin}^{ - 1}(sin2z)$

$\rm :\longmapsto\:y = 2z$

$\rm\implies \:y = 2 {tan}^{ - 1}5x$

On differentiating both sides w. r. t. x, we get

$\rm\implies \:\dfrac{d}{dx}y = \dfrac{d}{dx}2 {tan}^{ - 1}5x$

$\rm :\longmapsto\:\dfrac{dy}{dx} =2 \dfrac{d}{dx} {tan}^{ - 1}5x$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {tan}^{ - 1}x = \frac{1}{1 + {x}^{2} } \: }}}$

So, using this result, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 2 \times \dfrac{1}{1 + {(5x)}^{2} }\dfrac{d}{dx}5x$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{2}{1 + 25 {x}^{2} } \times 5$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{10}{1 + 25 {x}^{2} }$

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$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$