$Find \frac{dy}{dx}, if y = {(cosx)}^{x} + {(sinx)}^{\frac{1}{x}}$

Solution

$y = {(cosx)}^{x} + {(sinx)}^{\frac{1}{x}} For simplication, Let us consider y = A + B such that A = {(cosx)}^{x} and B = {(sinx)}^{\frac{1}{x}} Then, \frac{dy}{dx} = \frac{dA}{dx} + \frac{dB}{dx} . . . . . . . . . (i) A = {(cosx)}^{x}$

Taking logarithms on both the sides, we have,

log A = x log ( cosx )

$\frac{1}{A} \frac{dA}{dx} = \frac{d}{dx} [x \log (cosx)] \Rightarrow \frac{dA}{dx} = A \frac{d}{dx} [x \log (cosx)] = {(cosx)}^{x} [x \frac{d}{dx} [\log (cosx)] + \log (cosx) \frac{d}{dx} (x)] = {(cosx)}^{x} [x \frac{1}{\cos x} (- sinx) + \log (cosx) (1)] = {(cosx)}^{x} [- x tanx + \log (cosx)] . . . . . . . . (ii) Now, B = {(sinx)}^{\frac{1}{x}} B = {(sinx)}^{\frac{1}{x}}$

Taking logarithms on both the sides, we have,

$Log B = \frac{1}{x} \log (sinx) \frac{1}{B} \frac{dB}{dx} = \frac{d}{dx} [\frac{1}{x} \log (sinx)] \Rightarrow \frac{dB}{dx} = B \frac{d}{dx} [\frac{1}{x} \log (sinx)] = {(sinx)}^{\frac{1}{x}} [\frac{1}{x} \frac{d}{dx} (\log (sinx)) + \log (sinx) \frac{d}{dx} (\frac{1}{x})] = {(sinx)}^{\frac{1}{x}} [\frac{1}{x} \frac{1}{sinx} (cosx) + \log (sinx) (- \frac{1}{x^{2}})] = {(sinx)}^{\frac{1}{x}} [\frac{1}{x} cotx - (- \frac{1}{x^{2}}) \log (sinx)] = {(sinx)}^{\frac{1}{x}} [\frac{x cotx - \log (sinx)}{x^{2}}] . . . . . . . (iii)$

Now, on substituting (ii) and (iii) in (i) we get

$\frac{dy}{dx} = {(cosx)}^{x} [- x tanx + \log (cosx)] + {(sinx)}^{\frac{1}{x}} [\frac{x cotx - \log (sinx)}{x^{2}}]$

For Daily Free Study Material Join wiredfaculty

Continuity And Differentiability

$Find \frac{dy}{dx}, if y = {(cosx)}^{x} + {(sinx)}^{\frac{1}{x}}$

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Location

For Daily Free Study Material Join wiredfaculty

Continuity And Differentiability

Find dydx, if y = cosxx + sinx 1x

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

$Find \frac{dy}{dx}, if y = {(cosx)}^{x} + {(sinx)}^{\frac{1}{x}}$