How do you find the derivative of #(2e^(3x) + 2e^(-2x))^4#?

Question

Emma Aldridge · Accepted Answer

#4(2e^(3x)+2e^(-2x))^3(6e^(3x)-4e^(-2x))# #"differentiate using the "color(blue)"chain rule"# #"given "y=f(g(x))" then"# #dy/dx=f'(g(x))xxg'(x)larr" chain rule"# #f(g(x))=(2e^(3x)+2e^(-2x))^4# #rArrf'(g(x))=4(2e^(3x)+2e^(-2x))^3# #g(x)=2e^(3x)+2e^(-2x)# #rArrg'(x)=2e^(3x).d/dx(3x)+2e^(-2x).d/dx(-2x)# #color(white)(rArrg'(x))=6e^(3x)-4e^(-2x)# #rArrdy/dx=4(2e^(3x)+2e^(-2x))^3(6e^(3x)-4e^(-2x))#

Scarlett Allison · Answer

undefined To find the derivative of the given function $(2e^{3x} + 2e^{-2x})^4$, you can use the chain rule and the power rule for differentiation.

Let $u = 2e^{3x} + 2e^{-2x}$.

Now, differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = \frac{d}{dx}(2e^{3x} + 2e^{-2x})$$

Apply the chain rule and the derivative of $e^{ax}$ where $a$ is a constant:

$$\frac{du}{dx} = 2(3e^{3x}) - 2(2e^{-2x})$$

Simplify:

$$\frac{du}{dx} = 6e^{3x} - 4e^{-2x}$$

Now, apply the power rule for differentiation:

$$\frac{d}{dx}[(2e^{3x} + 2e^{-2x})^4] = 4(2e^{3x} + 2e^{-2x})^3 \cdot \frac{du}{dx}$$

Substitute the expression for $\frac{du}{dx}$ we found earlier:

$$\frac{d}{dx}[(2e^{3x} + 2e^{-2x})^4] = 4(2e^{3x} + 2e^{-2x})^3 \cdot (6e^{3x} - 4e^{-2x})$$

This is the derivative of the given function.