Question #99c7f

1 Answer
Oct 31, 2017

First derivative: 8e^(2x)secx+4e^(2x)secxtanx or 4e^(2x)secx(2+tanx)

Second derivative: 4e^(2x)(2+tanx)^2secx+4e^(2x)sec^3x

Explanation:

We'll begin by getting the first derivative.

[1]" "d/dx(4e^(2x)secx)

The first property we need is d/dx[cf(x)]=cd/dx[f(x)]. This means we can factor the constant out of the derivative (which in this case, is 4).

[2]" "=4d/dx(e^(2x)secx)

Next, we need the product rule: d/dx[f(x)g(x)=f'(x)g(x)+f(x)g'(x)]. In this case, let's have our f(x)=e^(2x) and our g(x)=secx.

[3]" "=4[(d/dxe^(2x))(secx)+(e^(2x))(d/dxsecx)]

The derivative of e^x is just e^x. But since we have e^(2x), we need to apply the chain rule. So d/dx(e^(2x))=2e^(2x).

[4]" "=4[(2e^(2x))(secx)+(e^(2x))(d/dxsecx)]

The derivative of secx is secxtanx.

[5]" "=4[(2e^(2x))(secx)+(e^(2x))(secxtanx)]

Simplifying, we get:

[6]" "=color(blue)(8e^(2x)secx+4e^(2x)secxtanx=4e^(2x)secx(2+tanx))

To get the second derivative, we'll get the derivative of the first derivative.

[1]" "d/dx[4e^(2x)secx(2+tanx)]

It's useful to note that 4e^(2x)secx is the same as our original equation.

Now we apply product rule.

[2]" "=[d/dx(4e^(2x)secx)][2+tanx]+(4e^(2x)secx)[d/dx(2+tanx)]

From earlier, we know that d/dx(4e^(2x)secx)=4e^(2x)secx(2+tanx). So we just plug that in.

[3]" "=[4e^(2x)secx(2+tanx)][2+tanx]+(4e^(2x)secx)[d/dx(2+tanx)]

The derivative of any constant is 0, and the derivative of tanx is sec^2x. So the derivative of 2+tanx is just sec^2x.

[4]" "=[4e^(2x)secx(2+tanx)][2+tanx]+[4e^(2x)secx][sec^2x]

We just simplify this now.

[5]" "=color(red)(4e^(2x)(2+tanx)^2secx+4e^(2x)sec^3x)