Question

How do you differentiate `f(x)=2secx+(2e^x)(tanx)`?

Answer 1

You use the sum rule and the product rule.

Explanation:

Take a look at your function

`f(x) = 2secx + 2e^x * tanx`

Notice that you can write this function as a sum of two other functions, let's say `g(x)` and `h(x)`, so that you have

`f(x) = g(x) + h(x)`

This will allow you to use the sum rule, which basically tells you that the derivative of a sum of two functions is equal to the sum of the derivatives of those two functions.

`color(blue)(d/dx(f(x)) = d/dx(g(x)) + d/dx(h(x))`

You also need to know that

`d/dx(tanx) = sec^2x`

and that

`d/dx(secx) = secx * tanx`

So, your function can be differentiate like this

`d/dx(f(x)) = d/dx(2secx) + d/dx(2e^x * tanx)`

`d/dx(f(x)) = 2 d/dx(secx) + d/dx(2e^x * tanx)`

For the derivative of `h(x)` you need to use the product rule, which tells you that the derivative of a product of two functions is equal to

`color(blue)(d/dx[a(x) * b(x)] = a^'(x)b(x) + a(x) * b^'(x))`

In your case, you have

`d/dx(2e^x * tanx) = d/dx(2 * e^x) * tanx + (2e^x) * d/dx(tanx)`

`d/dx(2e^x * tanx) = 2e^x * tanx + 2e^x * sec^2x`

Your original derivative will now be

`d/dx(f(x)) = 2 * secx * tanx + 2e^x tanx + 2e^x sec^2x`

`d/dx(f(x)) = color(green)(2 * [secx * tanx + e^x(tanx + sec^2x)])`