Question

Y=xtanx,show that xsin^2x(dy/dx)=x^2 tan x +y sin^2 x

Answer 1

$\textbf{Given:}$

$y=x\,tanx$

$\textbf{To prove:}$

$x\,sin^2x(\dfrac{dy}{dx})=x^2\,tan^2x+y\,sin^2x$

$\textbf{Solution:}$

$\text{Consider,}$

$y=x\,tanx$

$\text{Differentiate with respect to x}$

$\dfrac{dy}{dx}=x\,sec^2x+tanx\,1$

$\text{Multiply bothsides by x\,sin^2x }$

$x\,sin^2x\dfrac{dy}{dx}=x\,sin^2x(x\,sec^2x+tanx)$

$x\,sin^2x\dfrac{dy}{dx}=x^2\,sin^2x\,sec^2x+x\,sin^2x\,tanx$

$x\,sin^2x\dfrac{dy}{dx}=x^2\,sin^2x(\dfrac{1}{cos^2x})+(x\,tanx)\,sin^2x$

$x\,sin^2x\dfrac{dy}{dx}=x^2\,\dfrac{sin^2x}{cos^2x}+(x\,tanx)\,sin^2x$

$\implies\bf\,x\,sin^2x\dfrac{dy}{dx}=x^2\,tan^2x+y\,sin^2x$

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