How do you differentiate #y=sqrt(2-e^x)#?

Question

Paisley Johnson · Accepted Answer

Use the chain rule:
#(d(f(g(x))))/dx = (df(g))/(dg)(d(g(x)))/dx# Let #g(x) = 2 - e^x# then it follows that: #(d(g(x)))/dx= -e^x# #f(g) = sqrt(g)# and #(df(g))/(dg) = 1/(2sqrtg)# Substituting this into the chain rule: #(d(sqrt(2-e^x)))/dx =1/(2sqrtg)-e^x# #(d(sqrt(2-e^x)))/dx =-e^x/(2sqrtg)# Reverse the substitution for g: #(d(sqrt(2-e^x)))/dx =-e^x/(2sqrt(2-e^x))#

Luke Alvarez · Answer

undefined To differentiate $ y = \sqrt{2 - e^x} $, you can use the chain rule. The chain rule states that if you have a function inside another function, you differentiate the outer function first, then multiply it by the derivative of the inner function.

So, for $ y = \sqrt{2 - e^x} $, the outer function is the square root, and the inner function is $ 2 - e^x $.

The derivative of the outer function $ \sqrt{u} $ with respect to $ u $ is $ \frac{1}{2\sqrt{u}} $, and the derivative of the inner function $ u = 2 - e^x $ with respect to $ x $ is $ -e^x $.

Applying the chain rule, we get:

$$ \frac{dy}{dx} = \frac{1}{2\sqrt{2 - e^x}} \cdot (-e^x) $$

So, the derivative of $ y = \sqrt{2 - e^x} $ with respect to $ x $ is:

$$ \frac{dy}{dx} = \frac{-e^x}{2\sqrt{2 - e^x}} $$