Question

y=cos(xˣ)+(tanx)ˣ,Find dy/dx for the given function y wherever defined

Answer 1

Given, $y=cos(x^x)+(tanx)^x$

or, $y=cos(e^{xlogx})+e^{x.logtanx}$

now, differentiate with respect to x,

$\frac{dy}{dx}=\frac{d[cos(e^{xlogx})]}{dx}+\frac{d[e^{xlogtanx}]}{dx}$

$=-sin(e^{xlogx})\frac{d[e^{xlogx}]}{dx}+e^{xlogtanx}\frac{d[xlogtanx]}{dx}$

$=-sin(x^x).e^{xlogx}\frac{d(xlogx)}{dx}+(tanx)^x[x\frac{d[logtanx]}{dx}+logtanx\frac{dx}{dx}]$

$=-sin(x^x).x^x[x\times\frac{1}{x}+logx]+(tanx)^x[\frac{x}{tanx}sec^2x+logtanx]$

$=-sin(x^x)x^x[1+logx]+(tanx)^x[\frac{x.sec^2x}{tanx}+logtanx]$

Answer 2

HELLO DEAR,





GIVEN:-
Y = cos(x)ˣ + (tanx)ˣ

PUT U = cos(x)ˣ , V = tanx(x)ˣ

so, dY/dx = dU/dx + dV/dx


solving U = cos(x)ˣ
taking log both side,

logU = xlog(cosx)

(1/U)*dU/dx = -x.tanx + log(cosx)

dU/dx = cos(x)ˣ[log(cosx) - x.tanx]---------( 1 )

solving V = tan(x)ˣ

taking log both side,
logV = xlog(tanx)

(1/V)*dV/dx = x.{sec²x/tanx} + log(tanx)

dV/dx = tan(x)ˣ[log(tanx) + 2cosec2x]-------( 2 )


adding------( 1 ) & ------( 2 )

dU/dx + dV/dx = cos(x)ˣ[log(cosx) - x.tanx] + tan(x)ˣ[log(tanx) + 2cosec2x]

dy/dx = cos(x)ˣ[log(cosx) - x.tanx] + tan(x)ˣ[log(tanx) + 2cosec2x]


I HOPE ITS HELP YOU DEAR,
THANKS