Differentiate function;
J(y)={2y^2+[6y^3+(2y^4+3y^2)^6]^4}^2

Question

Differentiate function; 
J(y)={2y^2+[6y^3+(2y^4+3y^2)^6]^4}^2

zehanz · Answer

$J(y)=(2y^2+[6y^3+(2y^4+3y^2)^6]^4)^{2}$.

So a really nice looking function that will require the Chain Rule.

I am not into trying to split the function up in several smaller ones.
All we need to remember is that the Chain Rule tells us we must work form the outside to the inside. So we differentiate the last step (...)² so it becomes 2(...) and then multiply with the derivative of ..., and so on:

$J'(y)=2(2y^2+[6y^3+(2y^4+3y^2)^6]^4)\cdot derivative~of~stuff~between~brackets$.

zehanz · Answer

So this will be:

$J'(y)=2(2y^2+[6y^3+(2y^4+3y^2)^6]^4)\cdot (4y+4[6y^3+(2y^4+3y^2)^6]^3)\cdot ...$

Now there is more work to do, on the dots we must write the derivative of the function that is between the [] brackets:

$J'(y)=2(2y^2+[6y^3+(2y^4+3y^2)^6]^4)\cdot
\ (4y+4[6y^3+(2y^4+3y^2)^6]^3)\cdot \
(18y^2+6(2y^4+3y^2)^5)\cdot ...$

Now the last step can be done: we have to multiply with the derivative of $2y^4+3y^2$:

$J'(y)=2(2y^2+[6y^3+(2y^4+3y^2)^6]^4)\cdot
\ (4y+4[6y^3+(2y^4+3y^2)^6]^3)\cdot \
(18y^2+6(2y^4+3y^2)^5)\cdot (8y^3+6y)$.

I must say I pity you. This is pure torture! It is appearently meant to check your understanding the Chain Rule, but with a function this ugly, the possibility of making a tiny error along the way is rather big. If so, it is almost impossible to see where the fault is, so you have to start all over again.

I definitely hope I didn't make one.

anonymous · Answer

@ZeHanz,thanks a lot for the help I appreciate the detail explanation.