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

1 Answer
Jul 29, 2015

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)])#