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$$y = {\tan{4 \sin^2(5 e+9)\over e^5+\tan 33°}}+{\sqrt[7]{\ln {{\sin 2x - \cos (e-x^2)}\over \cos^33x}}}+e^{5-9x \ln e}$$

In the end I need to have $y'$

Like honestly I don't even know where to start..?

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    Honestly, I would never even begin to differentiate that horrid looking thing by hand. If you must, you can note that the first term does not depend on $x$ and hence vanishes after differentiation with respect to $x$. To differentiate the other terms the easiest is probably multiple uses of the chain rule. – 2017-01-03
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    University kills me... – 2017-01-03
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    Thanks thats what I needed. Should I delete the post? – 2017-01-03
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    No, don't delete the question. Others can learn from it, and someone may also give an answer that helps more. – 2017-01-03
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    Chain rule is your friend. It is not too hard, just start differentiating $(f)^{1/7}$. – 2017-01-03
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    This is *not* a hard problem. You are just intimidated by its appearance. The first term is constant, and the last term is a constant times $e^{-9x}$, so only the middle term needs work. For starters, it's of the form $(\log f(x))^c$, so work with that... – 2017-01-03

3 Answers 3

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First of all, notice the first term is a constant. Then, simplify things (e.g. lne=1) and use the rules for differentiation, for example the chain rule. You should ideally also care about finding the domain (i.e. where y=y(x) is differentiable).

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HINT: simbolically the mid term can be written as

$$(g\circ h)^c$$

where $c=1/7$, $g=\ln$ and $h=\frac{\sin (f_1)-\cos(f_2)}{(\cos (f_3))^3}$. And you know that

$$[(f)^c]'= c(f)^{c-1}f',\quad\text{and}\quad (g\circ h)'=(g'\circ h)h'$$

and

$$[\ln(f)]'=\frac{f'}{f},\quad [\sin(f)]'=f'\cos(f),\quad\text{and}\quad[\cos(f)]'=-f'\sin(f)$$

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As said in comments and answers, only the middle term is problematic $$A={\sqrt[7]{\ln {{\sin (2x) - \cos (e-x^2)}\over \cos^3(3x)}}}\implies A^7={\ln {{\sin (2x) - \cos (e-x^2)}\over \cos^3(3x)}}$$ $$e^{A^7}={{\sin (2x) - \cos (e-x^2)}\over \cos^3(3x)}\implies 7A^6 e^{A^7} A'=\left({{\sin (2x) - \cos (e-x^2)}\over \cos^3(3x)}\right)'$$