Question

if y = tanx . cos^2x then dy/dx will be

Answer 1

If y = tanx . cos²x then dy/dx = cos 2x

Given :

y = tanx . cos²x

To find :

The value of dy/dx

Solution :

Step 1 of 2 :

Write down the given function

Here the given function is

$\displaystyle \sf{y = tanx. {cos}^{2}x }$

$\displaystyle \sf{ \implies y = \frac{sinx}{cosx} . {cos}^{2}x}$

$\displaystyle \sf{ \implies y = sinx.cosx}$

$\displaystyle \sf{ \implies y = \frac{1}{2} \times 2 sinx.cosx}$

$\displaystyle \sf{ \implies y = \frac{1}{2} sin2x}$

Step 2 of 2 :

Find the value of dy/dx

$\displaystyle \sf{ y = \frac{1}{2} sin2x}$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \implies \frac{dy}{dx} = \frac{d}{dx} \bigg( \frac{1}{2} sin2x \bigg)}$

$\displaystyle \sf{ \implies \frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} \bigg( sin2x \bigg)}$

$\displaystyle \sf{ \implies \frac{dy}{dx} = \frac{1}{2} \times cos2x \times \frac{d}{dx} \bigg( 2x \bigg)}$

$\displaystyle \sf{ \implies \frac{dy}{dx} = \frac{1}{2} \times cos2x \times 2\frac{d}{dx} \bigg( x \bigg)}$

$\displaystyle \sf{ \implies \frac{dy}{dx} = \frac{1}{2} \times cos2x \times 2 \times 1}$

$\displaystyle \sf{ \implies \frac{dy}{dx} = cos \: 2x }$

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Answer 2

Answer:

1-2sin^2x

Refer the attachment