Question

How do you differentiate `g(x) = 3xsin^2(4x) + secx`?

Answer 1

Make use of the Chain Rule and Product Rule. `(dg)/dx = 3sin^2(4x) + 24xsin(4x)cos(4x)+tan(x)sec(x)`

Explanation:

Our first term will have to have the product rule applied. The product rule states that given `f(x)=g(x)h(x), f'(x) = g'(x)h(x) + g(x)h'(x)`

Thus our first term...

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

The derivative for this second term will require use of the chain rule, and of substitution. If we declare `u(x) = sin(4x)`, then we have `(du)/dx = 4 cos(4x) sin^2(4x) = u^2`, meaning `d/dx u^2 = d/du (u^2) * (du)/dx = 2u*(du)/dx. = 2 sin(4x) * 4cos(4x) = 8sin(4x)cos(4x)`

This means:

`d/dx (3xsin^2(4x)) = 3sin^2(4x) + 3x(8 sin(4x)cos(4x)) = 3sin^2(4x) + 24xsin(4x)cos(4x)`

Meanwhile, the derivative of `sec(x)` can be found using the quotient rule and the definition of the secant.

`sec(x) = 1/cos(x) -> d/dx (1/cos x) = ((0*cosx) - (-sin x))/(cos^2x) = sin x / (cos^2x) = tan(x)* 1/cos(x) = tanx secx`

Thus we have...

`(dg)/dx = 3sin^2(4x) + 24xsin(4x)cos(4x)+tan(x)sec(x)`