Consider this way of differentiating...I have never seen it done...but it is an adaptation of logarithimic differentiating so I am sure other people have done it

It goes something like this...say you have $\displaystyle y=f(u(x))$....now say that $\displaystyle f(x)$ is difficult to integrate....why not then do this

[math\f^{-1}(y)=u(x)[/tex]

For example this is a way to either be a smart*** on a test if you forget or it is a way I have discovered of proving a trigonometric identity

$\displaystyle y=sin^{-1}(x)$......say you forget that derivative...well do this

$\displaystyle \sin(y)=x$...differentiating we see that

$\displaystyle y'\cdot\cos(y)=1\Rightarrow{y'=\frac{1}{cos(y)}}\R ightarrow{y'=\frac{1}{\cos(\sin^{-1}(x))}}$

The trig proof comes from the fact taht $\displaystyle y=sin^{-1}(x)\Rightarrow{y'=\frac{1}{\sqrt{1-x^2}}}$

Now since $\displaystyle y'=y'\Rightarrow{\frac{1}{\cos(\sin^{-1}(x))}=\frac{1}{\sqrt{1-x^2}}}\Rightarrow{\cos(\sin^{-1}(x))=\sqrt{1-x^2}}$

which is a famous identity...this can be done for most things

Another example would be $\displaystyle y=\ln(u(x))$

Forget how to take the derivative of ln(x)? Do this...

$\displaystyle e^{y}=u(x)\Rightarrow{y'\cdot{e^{y}}=u'(x)}\Righta rrow{y'=\frac{u'(x)}{e^{y}}}$

Now making the sub$\displaystyle y=\ln(u(x))$ we get

$\displaystyle y'=\frac{u'(x)}{e^{\ln(u(x))}}=\frac{u'(x)}{u(x)}$

which is the derivative of ln((u(x))