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Find the derivatives of the given functions.$$y=\frac{7 \ln 3 x}{e^{2 x}+8}$$

$\frac{d y}{d x}=\frac{\left(e^{2 x}+8\right)-14 x e^{2 x} \ln 3 x}{x\left(c^{2 x}+8\right)^{2}}$

Calculus 1 / AB

Chapter 27

Differentiation of Transcendental Functions

Section 6

Derivative of the Exponential Function

Derivatives

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All right. We want to find dy dx for the function. Why is it the seven times the natural longer than the three X over either the two X plus eight to do so we're gonna use derivatives shortcuts. We picked up through a single variable kelp primarily rule 0 to 3 listed here on left. Rule zeros is D D X BV you is B V u L n B D X D dx. Either you Izzy the U D X rules +123 other powerful product rule in chain, respectively. First, let's rewrite why and more easily differential terms. So we have Y equals seven Ln three X times either to expose 18 negative first. Now we see how the chain will and product rule or more essential to this problem. So we have to take seven Ln three X. And even the to expose eight negative first derivative separately. For the product rule. That's we have dy dx equals on the left. Seven times 3/3 X. Either to expose 18 91st. This is because Ellen three X derivative 3/3 X on the right. We have seven Ln three X times negative 92 X plus 8 to 92nd times either two x times two applying rule zero. And the chain rule that we have final solution highlighted the bottom of the screen. When we multiply the first term by either to expose it over either to expose eight to combine the denominators. We also multiply on the right by X over X to obtain final solution seven times E to the to expose eight minus 14 X. Ln three X. Either two X divided by X times two X eight squared.

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