Question

differentiate chapter tan inverse (log x)​

Answer 1

Answer:

let, $y=tan^{-1} (logx)$

diff...............wrt"x"

$\frac{dy}{dx} =\frac{d(tan^{-1}(logx))}{dx}$

,$\frac{dy}{dx} =\frac{1}{1+(logx)^{2} } *\frac{d(logx)}{dx}$

,$\frac{dy}{dx} =\frac{1}{1+(logx)^{2} } *\frac{1}{x} \\\\\frac{dy}{dx} =(\frac{1}{1+(logx)^{2} } )\frac{1}{x}$

Answer 2

Answer:

dy/dx=1 / x(1+logx)

Step-by-step explanation:

let y=tan ⁻¹(logx)

and t= logx

dt/dx=1/x

y=tan ⁻¹(t)

dy/dt =1 / (1+t²)

now dy/dx=dy/dt*dt/dx

= 1 / (1+t²) *1/x

putting t=logx

dy/dx=1 / x(1+logx)