I came across some exponential and logarithm practice problems at liavas.net and made my way through all of them and came to the final question 3. Find the derivative of $y=\cosh(x^2)$ And so I proceeded with the chain rule $$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}$$ And now we let $y=\cosh(u)$ where $u=x^2$ therefore $\frac{dy}{du}=\sinh(u)$ and $\frac{du}{dx}=2x$ And so $\frac{dy}{dx}=2x\sinh(x^2)$ no? The solution shown has $\frac{dy}{dx}=2x\sinh(x)$ and I was wondering where I've gone wrong.
$\endgroup$ 4 Reset to defaultDifferentiate $\cosh(x^2)$
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