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I have an assignment in which I have to differentiate a function that employs both exponential and hyperbolic trig parts. The equation is:

$$ y=\frac{e^{4x}}{x^3 \cosh (2x)} $$

I know I need to use a combination of Quotient Rule, Chain Rule and Product Rule, and i get as far as

$$ y'=\frac{e^{4x} [x\cosh(2x)-2 \cosh⁡(2x)-2x \sinh(2x)]}{x^4\cosh^2(2x)} $$

I don't even know if what I have done so far is correct.

I know I can use Logarithmic Differentiation for this problem as well, but I don't get anywhere near the required answer using it, so some advice on that would be awesome too.


The answer I need to get is

$$ y'=\frac{e^{4x} [(4x-3)\cosh(2x)+2x\sinh(2x)]}{x^4} $$

I have no idea where the $(4x-3)$ comes from, or where the $\cosh^2(2x)$ disappears to from the denominator

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Logarithmic differentiation seems like a good plan; anything to avoid a combined quotient and product rule.

$$ \ln(y) = 4x - 3 \ln x - \ln[\cosh(2x)]\\ \frac{y'}{y} = 4 - \frac 3x - \frac{2\sinh(2x)}{\cosh(2x)} = \frac{4 x \cosh(2x) - 3 \cosh(2x) - 2x \sinh(2x)}{x \cosh(2x)} =\\ \frac{(4x - 3)\cosh(2x) - 2x \sinh(2x)}{x \cosh(2x)} $$ So, the answer we should end up with is $$ y' = \frac{e^{4x}[(4x - 3)\cosh(2x) - 2x \sinh(2x)]}{x^4 \cosh^2(2x)} $$ It seems that they have erroneously left out the $\cosh^2(2x)$ in the denominator.

Your mistake in differentiating is that you forgot to apply the chain rule when differentiating $e^{4x}$.

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  • $\begingroup$ Thank you. I retried and got it. Awesome answer. Wish I could give more than one upvote. $\endgroup$ – Rene Aug 4 '16 at 19:12

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