How do you differentiate # (e^(2x) + 2x) ^0.5#?

Question

Ezekiel Alexander · Accepted Answer

#(e^(2x) + 1)/(e^(2x) + 2x )^(1/2) #

#f(x) = (e^(2x )+ 2x )^(1/2)#

using the chain rule :(twice)

#f'(x) = 1/2(e^(2x) + 2x )^(-1/2) . d/dx (e^(2x) + 2x )#

#= 1/2(e^(2x) + 2x )^(-1/2) (e^(2x) .d/dx(2x) + 2 )#

#= 1/2(e^(2x) +2x )^(-1/2) (2e^(2x) + 2 ) #

#= 1/2 (e^(2x) +2x )^(-1/2) .2 ( e^(2x) + 1 ) #

#rArrf'(x) = (e^(2x) + 1 )/(e^(2x) + 2x )^(1/2) #

Ezekiel Alexander · Answer

undefined To differentiate (e^(2x) + 2x)^0.5, you can use the chain rule.

1. Let $ u = e^{2x} + 2x $.
2. Rewrite the expression as $ u^{0.5} $.
3. Apply the power rule: $ \frac{d}{du} u^{0.5} = \frac{1}{2}u^{-0.5} $.
4. Apply the chain rule: $ \frac{d}{dx} (e^{2x} + 2x)^{0.5} = \frac{1}{2}(e^{2x} + 2x)^{-0.5} \cdot \frac{d}{dx}(e^{2x} + 2x) $.

5. Differentiate $ e^{2x} + 2x $ with respect to $ x $:
   $ \frac{d}{dx}(e^{2x} + 2x) = 2e^{2x} + 2 $.

6. Substitute back in:
   $ \frac{d}{dx} (e^{2x} + 2x)^{0.5} = \frac{1}{2}(e^{2x} + 2x)^{-0.5} \cdot (2e^{2x} + 2) $.

7. Simplify if necessary.

Adam Adkins · Answer

undefined To differentiate (e^(2x) + 2x) ^0.5, you can use the chain rule. The derivative of a function raised to a power can be calculated using the chain rule. Let's denote the given function as u(x) = (e^(2x) + 2x) and v(x) = 0.5. Then, the function can be rewritten as y = u(x) ^ v(x). Now, using the chain rule, the derivative dy/dx is given by dy/dx = v(x) * u(x)^(v(x)-1) * u'(x) + ln(u(x)) * u(x) ^ v(x) * v'(x). Substituting the values of u(x) and v(x), and their derivatives, we get dy/dx = 0.5 * (e^(2x) + 2x)^(-0.5) * (2e^(2x) + 2) + ln(e^(2x) + 2x) * (e^(2x) + 2x)^(0.5) * 0. Applying algebraic simplification, the final expression for the derivative dy/dx is dy/dx = (e^(2x) + 1) / (2(e^(2x) + 2x)^(0.5)).