Question

d/dx (x+1/x+log x+tanx)​

Answer 1

ANSWER:

• Derivative of x + 1/x + log x + Tan x = 1 - 1/x² + 1/x + sec² x.

GIVEN:

• x + 1/x + log x + Tan x.

TO FIND:

• Derivative of x + 1/x + log x + Tan x.

EXPLANATION:

Let y = x + 1/x + log x + Tan x

Let A = x, B = 1/x, C = log x and D = Tan x

y = A + B + C + D

Take A:

A = x

Differentiate w.r.t x

$\sf \dfrac{dA}{dx} = \dfrac{d}{dx}x$

$\boxed{\bold{\large{\gray{\dfrac{d}{dx}x = 1}}}}$

$\sf \dfrac{dA}{dx} = 1$

Take B:

B = 1/x

Differentiate w.r.t x

$\sf \dfrac{dB}{dx} = \dfrac{d}{dx}x^{-1}$

$\boxed{\bold{\large{\gray{\dfrac{d}{dx} {x}^{n} = n{x}^{n - 1} }}}}$

$\sf \dfrac{dB}{dx} = - 1 (x^{-1 - 1})$

$\sf \dfrac{dB}{dx} = - 1 (x^{-2})$

$\sf \dfrac{dB}{dx} = - \dfrac{1} {x^{2}}$

Take C:

C = log x

Differentiate w.r.t x

$\sf \dfrac{dC}{dx} = \dfrac{d}{dx}log\ x$

$\boxed{\bold{\large{\gray{\dfrac{d}{dx}log\ x = \dfrac{1}{x} }}}}$

$\sf \dfrac{dC}{dx} = \dfrac{1}{ x}$

Take D:

D = Tan x

Differentiate w.r.t x

$\sf \dfrac{dD}{dx} = \dfrac{d}{dx}Tan\ x$

$\boxed{\bold{\large{\gray{\dfrac{d}{dx}Tan\ x = {Sec}^{2} \ x }}}}$

$\sf \dfrac{dD}{dx} = {Sec}^{2} \ x$

$\sf We \ know\ that \ y = A + B + C + D$

Differentiate w.r.t x

$\sf \dfrac{dy}{dx} =\dfrac{dA}{dx} +\dfrac{d B}{dx} + \dfrac{dC}{dx} + \dfrac{dD}{dx}$

Substitute the respective values.

$\sf \dfrac{dy}{dx} =1 - \dfrac{1} {x^{2}} + \dfrac{1}{ x} + {Sec}^{2} \ x$

Hence the derivative of x + 1/x + log x + Tan x = 1 - 1/x² + 1/x + sec² x.