Derivative of $\;\displaystyle f(x) = \sinh \left(\cosh \left(x^9\right)\right) \,$ ?

Okay, so I tried u substitution:

$ u = \cosh(x^9)$

$ du = \sinh(x^9) \cdot 9x^8 dx$


$$ \sinh(u) \frac{du}{\sinh(x^9) 9x^8}$$

As you can see, I have the two variables, u and x.

I'm guessing I'm supposed to know how to do double substitution, another one for x^9, but I don't know how to do it for derivatives... so I feel like there's a gap in my math knowledge, something I was supposed to know up until this point... So if you'd just go ahead and show me how to do that, that'd be great.


$f'(g(x)) \cdot g'(x)$

So what's my $f'$? well, it's sinh(something), anyhow, derivative of sinh(something) = cosh(something)

then I just plug g(x) into the derivative of f

so my $f'(g(x)) \cdot g'(x) = \cosh(\cosh(x^9)) \cdot \sinh(x^9) 9x^8$



Use the chain-rule and write

$$\frac{d}{dx}\sinh (\cosh (x^9))=\left(\left.\frac{d\,\sinh (u)}{d u}\right)\right|_{u=\cosh (x^9)}\times\left(\left.\frac{d\,\cosh (v)}{d v}\right)\right|_{v=x^9}\times\left(\frac{d\,x^9}{dx}\right)$$

  • $\begingroup$ if you do u substitution $u = cosh(x^9)$ then where's your du? With your du you also will need to divide by $ \sinh(x^9) \cdot 9x^8 dx$ just like I had stated above. $\endgroup$ – Jack Sep 28 '15 at 3:29
  • $\begingroup$ @jack I assure you, this answer is correct. Note the multiplication by the next two terms. $\endgroup$ – Mark Viola Sep 28 '15 at 13:48

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