Fundamental derivatives

1 Oct 2019 Leave a Comment

What it takes to be able to calculate any derivative is to know the rules of differentiation and the fundamental derivatives, the ones you have to memorize. If you are reading this, you have probably learnt the rules, which I covered in my previous lesson. If you prefer, you can study this part first and then go to the rules.

This lesson is going to be shorter because there isn’t much you need to understand, you just have to memorise the following derivatives:

$\frac{\mathrm{d} }{\mathrm{d} x}\ln(x)=\frac{1}{x}$

Example:

$\frac{\mathrm{d} }{\mathrm{d} x}\ln(x^4)=\frac{1}{x^4}4x^3=\frac{4}{x}$

Note: the chain rule was applied here.

$\frac{\mathrm{d} }{\mathrm{d} x}e^x=e^x$

Example:

$\frac{\mathrm{d} }{\mathrm{d} x}e^{x^2-7x}=(2x-7)e^{x^2-7x}$

Chain rule applied. It will be like this for every example.

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)$

Example:

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(x^7-4)=\cos(x^7-4)7x^6=7x^6\cos(x^7-4)$

$\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-\sin(x)$

Example:

$\frac{\mathrm{d} }{\mathrm{d} x}\cos(x^3+2x^2-3)=-\sin(x^3+2x^2-3)(3x^2+4x)=$ $-(3x^2+4x)\sin(x^3+2x^2-3)$

$\frac{\mathrm{d} }{\mathrm{d} x}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$

Example:

$\frac{\mathrm{d} }{\mathrm{d} x}\arcsin(2x^3)=\frac{1}{\sqrt{1-{(2x^3)}^2}}6x^2$

$\frac{\mathrm{d} }{\mathrm{d} x}\arccos(x)=-\frac{1}{\sqrt{1-x^2}}$

Example:

$\frac{\mathrm{d} }{\mathrm{d} x}\arccos(12x^3-4x+5)=-\frac{1}{\sqrt{1-{(12x^3-4x+5)}^2}}{(36x^2-4)}$

$\frac{\mathrm{d} }{\mathrm{d} x}\arctan(x)=\frac{1}{x^2+1}$

Example:

$\frac{\mathrm{d} }{\mathrm{d} x}\arctan(5x^{-3})=\frac{1}{{(5x^{-3})}^2+1}{(-15x^{-4})}$

Now that you know these derivatives and the rules, you can calculate the derivative of nested functions! Here are some examples:

$\frac{\mathrm{d} }{\mathrm{d} x}\ln(\cos(x))=\frac{1}{\cos(x)}(-\sin(x))=-\tan(x)$ $\frac{\mathrm{d} }{\mathrm{d} x}\sin(\cos(x))=\cos(\cos(x))(-\sin(x))=-\sin(x)\cos(\cos(x))$ $\frac{\mathrm{d} }{\mathrm{d} x}\sin(\ln(x))=\cos(\ln(x))\frac{1}{x}$ $\frac{\mathrm{d} }{\mathrm{d} x}\ln(\sin(7x^5))=\frac{1}{\sin(7x^5)}\cos(7x^5)35x^4$ $\frac{\mathrm{d} }{\mathrm{d} x}\ln(\arccos(x^2+4x))=\frac{1}{\arccos(x^2+4x)}(-\frac{1}{\sqrt{1-(x^2+4)^2}})(2x+4)$

And now, a monster (try this out yourself first!):

$\frac{\mathrm{d} }{\mathrm{d} x}\arctan(\ln(\sqrt{\arcsin(\sqrt[5]{\sin(12x^{-4})})}\ln(\cos(\arccos(x^3)))))$ $\frac{\mathrm{d} }{\mathrm{d} x}\arctan(\ln(\sqrt{\arcsin(\sqrt[5]{\sin(12x^{-4}e^{-x^3})})}\ln(\cos(\arccos(x^3)))))=\frac{1}{{(\ln(\sqrt{\arcsin(\sqrt[5]{\sin(12x^{-4}e^{-x^3})})}\ln(\sin(\arccos(x^3)))))}^2}\frac{1}{\sqrt{\arcsin(\sqrt[5]{\sin(12x^{-4}e^{-x^3})})}\ln(\sin(\arccos(x^3)))}((\frac{1}{2}\arcsin{(\sqrt[5]{\sin(12x^{-4}e^{-x^3})}))}^{-\frac{1}{2}}\frac{1}{1-{(\sqrt[5]{\sin(12x^{-4}e^{-x^3})})}^2}(\frac{1}{5}\sin(12x^{-4}e^{-x^3}))^{-\frac{4}{5}}cos(12x^{-4}e^{-x^3})(-48x^{-5}e^{-x^3}+12x^{-4}(-3x^{-4}e^{-x^3}))\ln(\sin(\arccos(x^3)))+(\sqrt{\arcsin(\sqrt[5]{\sin(12x^{-4}e^{-x^3})})}\frac{1}{\sin(\arccos(x^3))}\cos(\arccos(x^3))(-\frac{1}{\sqrt{1-(x^3)^2)}})3x^2)$

Just kidding! I mean, if you really want to give it a try, do it, but take your time. It’s a very long procedure (I am not even 100% sure this is the correct result, and keep in mind that I didn’t perform any simplification. I just applied the rules and wrote down the derivatives for each step. You can try this one instead:

$\frac{\mathrm{d} }{\mathrm{d} x}\cos(\frac{1}{\ln(x)})e^{-x}$

Here’s the procedure:

$\frac{\mathrm{d} }{\mathrm{d} x}\cos(\frac{1}{\ln(x)})e^{-x}=-\sin(\frac{1}{\ln(x)})(-\frac{\frac{1}{x}}{\ln^2(x)})e^{-x}+\cos(\frac{1}{\ln(x)})(-e^{-x})$

The reciprocal, exponential and chain rules were used.

If you have any doubt, something was not clear, or you need help to solve a derivative, leave a comment and I’ll reply asap!

In the next lesson I will talk about the rules of integration. Stay tuned!

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Differentiation Calculus, fundamental derivatives

Add just a bit of pi

Fundamental derivatives

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Fundamental derivatives

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