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

Question

Luke Alvarez · Accepted Answer

You use the sum rule and the product rule.

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)])#

Cameron Alexander · Answer

undefined To differentiate $f(x) = 2\sec(x) + (2e^x)(\tan(x))$, use the sum rule, which allows us to differentiate each term separately, and the product rule for the second term, since it is the product of two functions. The derivative of $2\sec(x)$ and $2e^x \tan(x)$ are found as follows:

1. **Differentiate $2\sec(x)$:**
   - The derivative of $\sec(x)$ is $\sec(x)\tan(x)$.
   - Thus, the derivative of $2\sec(x)$ is $2\sec(x)\tan(x)$.

2. **Differentiate $2e^x \tan(x)$:**
   - This requires the product rule: If $u(x) = 2e^x$ and $v(x) = \tan(x)$, then $u'(x) = 2e^x$ (since the derivative of $e^x$ is $e^x$) and $v'(x) = \sec^2(x)$ (the derivative of $\tan(x)$).
   - Apply the product rule: $(uv)' = u'v + uv'$.
   - So, the derivative of $2e^x \tan(x)$ is $2e^x \tan(x) + 2e^x \sec^2(x)$.

Putting it all together:

$$f'(x) = 2\sec(x)\tan(x) + 2e^x \tan(x) + 2e^x \sec^2(x)$$

This expression is the derivative of $f(x) = 2\sec(x) + (2e^x)(\tan(x))$.