Question

Differentiate w.r.t.x. : cos² (log (2x + 3))

Answer 1

d/dx cos^2 (log (2x+3))

(d/dx x^2= 2x)

=2cos (log(2x+3)) d/dx log( 2x+3)

= 2cos(log(2x+3)) 1/(2x+3)

=2 cos(log(2x+3)/(2x+3)

Answer 2

Answer:

$2 cos(log(2x+3).(-sin(log(2x+3)).(\frac{2}{2x+3} )\\ \\ or\\ \\ \frac{-2sin[2log(2x+3)]}{2x+3}$

Step-by-step explanation:

apply chain rule for differentiating the given function

Formula used:

$\frac{df(x)^{2} }{dx} =2f(x).\frac{df(x)}{dx}\\ \\ \frac{d cos x}{dx} =-sin x\\ \\ \frac{d logf(x)}{dx} =\frac{1}{f(x)} .\frac{df(x)}{dx}$

So here

$\frac{cos^{2}[log(2x+3) }{dx} = 2cos[log(2x+3)].\frac{cos[log(2x+3)]}{dx} \\ \\ \\ =2cos[log(2x+3)].(-sin[log(2x+3)].\frac{d log(2x+3)}{dx} \\ \\ \\ =2cos[log(2x+3)].(-sin[log(2x+3)].(\frac{1}{2x+3} )\frac{d(2x+3)}{dx} \\ \\ \\=2cos[log(2x+3)].(-sin[log(2x+3)]).\frac{2}{2x+3}$

now we can simplify the expression

as we know that 2 sin x cos x = sin 2x

So,

$\frac{-2sin[2log(2x+3)]}{2x+3}$