Question

If x= e^tan-1 y-x²/x², then dy/dx=........,Select Proper option from the given options.
(a) 2x(tan(log x)+1)
(b) 2x(tan(log x)+1)+x²sec(logx)
(c) 2x(tan(log x)+1)+x²sec(log x)
(d) 0

Answer 1

Given, $x=e^{tan^{-1}\frac{y-x^2}{x^2}}$

$logx=tan^{-1}\frac{y-x^2}{x^2}$

$tanlogx=\frac{y-x^2}{x^2}$

$x^2tanlogx =y-x^2$

$x^2(1 + tanlogx)= y$

now, differentiate y with respect to x,

$\frac{dy}{dx}=\frac{d\{x^2(1+tanlogx)\}}{dx}$

$\qquad=x^2\frac{d(1+tanlogx)}{dx}+(1+tanlogx)\frac{dx^2}{dx}$

$\qquad=x^2sec^2logx\frac{d(logx)}{dx}+(1+tanlogx).2x$

$\qquad=x^2sec^2logx\frac{1}{x}+2x(1+tanlogx)$

$\qquad=2x(1+tanlogx)+xsec^2(logx)$

here little mistake in option (b) and (c) , both are same. one of them must be 2x(1 + tanlogx) + xsec^2(logx)