Question

y^2 cos(1/x)=a^2 find dy/dx​

Answer 1

Answer:

Refer to the attachment.

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Answer 2

Differentiation is a process of finding the derivative or rate of change of one quantity with respect to another.

• It is denoted by $\frac{dy}{dx}$ or f'

The basic formula to find the differentiation of a polynomial function (say) $y=x^n$ is,

                       $\frac{dy}{dx}=\frac{d(x^n)}{dx}=nx^{n-1}$ This is called 'the power rule'.

                         

Given  ,  $y^2 cos(\frac{1}{x})=a^2$ , let us find $\frac{dy}{dx}$

Since we have to find differentiation of y with respect to x let us solve the given equation for 'y'.

                      $y^2 cos(\frac{1}{x})=a^2\\\\y^2 =\frac{a^2}{cos(\frac{1}{x})}\\\\y=\sqrt{\frac{a^2}{cos(\frac{1}{x})}}$

                   $\frac{dy}{dx}=\frac{d}{dx}(\sqrt{\frac{a^2}{cos(\frac{1}{x})}}})$  - - - - (1)

From trigonometric ratios we have,  

                 $\frac{1}{cos(\frac{1}{x})} = sec(\frac{1}{x})$

∴ equation (1) can be written as,

                    $\frac{dy}{dx}=\frac{d}{dx}(\sqrt{a^2sec(\frac{1}{x})})\\\\\frac{dy}{dx}=a\,\frac{d}{dx}(\sqrt{sec(\frac{1}{x})})$ - - - - (2)

• Trigonometric functions are differentiable, and the differentiation of secx is

                  $\frac{d(secx)}{dx}=secx\,\,tanx$

• and       $\frac{d(\sqrt{x}\,)}{dx} = \frac{1}{2\sqrt{x}}$
• using power rule to differentiate $\frac{1}{x}$

         $\frac{d(\frac{1}{x})}{dx}=\frac{d(x^{-1})}{dx} = -x^{-2} \\\\\frac{d(\frac{1}{x})}{dx}= -\frac{1}{2}$

Let us solve (2) using the points mentioned above,

                  $\frac{dy}{dx}=a\,\frac{d}{dx}(\sqrt{sec(\frac{1}{x})}) \\\\ \frac{dy}{dx}=a\,\frac{1}{2\sqrt{sec(\frac{1}{x})}}\frac{d}{dx}({sec(\frac{1}{x})})$                 (Using chain rule)

                 $\frac{dy}{dx}=-\frac{a}{2x^2}\,\frac{1}{\sqrt{sec(\frac{1}{x})}}\,sec(\frac{1}{x})}\,\,tan(\frac{1}{x})}\\\\\frac{dy}{dx}=-\frac{a\,\,\sqrt{sec(\frac{1}{x})}\,\,tan(\frac{1}{x})}{2x^2}$

∴    $\bf \frac{dy}{dx}=-\frac{a\,\,\sqrt{sec(\frac{1}{x})}\,\,tan(\frac{1}{x})}{2x^2}$  is the required solution.

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