How to differentiate this?

it's clear you have a quotient. so use the quotient rule for differentiating.

Let the numerator be called u(x) and the denominator be v(x). Find the first derivative of u(x) and of v(x).

Then apply the quotient rule.

dy/dx = {v(x)[first derivative of u] - u(x)[first derivative of v] } / v^2

sin(x)/cos(x)=tan(x)

d/dx(tan(x)) = sec^2(x)

and

using the product rule (d(f(x)*g(x))/dx=f'(x)*g(x)+f(x)*g'(x))

d(x^2)/dx=2x

d(sin(4x+3))/dx = 4cos(4x+3)

hence:

d/dx(x^2 sin(4 x+3)) = 2 x (sin(4 x+3)+2 x cos(4 x+3))

sin x / cos x = tan x

d/dx (tan x) = sec^2 x

alternative method:

y = sin x (cos x)^(-1)

apply the product rule

y' = sin x d/dx ( cos x)^(-1) + (cos x)^(-1) d/dx (sin x)

y' = sin x (-1) (cos x)^(-2) (-sin x) + (cos x)^(-1) cos x

y' = sin^2 x / cos^2 x + cos x / cos x

y' = tan^2 x + 1

y' = sec^2 x

-------------------------------------

y= x^2 sin(4x+3)

apply the product rule

y' = x^2 d/dx ( sin(4x+3) ) + sin (4x+3) d/dx (x^2)

y' = x^2 cos(4x+3) d/dx (4x+3) + sin(4x+3) (2x)

y' = x^2 cos(4x+3) (4) + (2x) sin(4x+3)

y' = 4x^2 cos(4x+3) + 2x sin(4x+3)