Question

Find the derivative..

Answer 1

\[y = (x^{2}+1)\sqrt{x^{3}-2x+3}\]

Answer 2

@ganeshie8 @Preetha @AravindG

Answer 3

it may help you to write it as y = A.B where

A = x^2 + 1
B = (x^3 - 2x + 3)^(1/2)

so dy/dx = dA/dx.B + A.dB/dx

you can de-compose B further if that is still looking tricky
let B = C^(1/2)
where C = x^3 - 2x + 3
thus dB/dx = dB/dC.dC/dx

that's all there is to it apart from the basic start point, that d/dx( x^n) = nx^(n-1).

eventually you will be doing all of this in your head...

Answer 4

Thanks a bunch @IrishBoy123 :)

Answer 5

Find y':

\[y = \frac{ 1 }{ ((3x + 1)^{4} }\]

Answer 6

@UnkleRhaukus

Answer 7

this one is a lot easier.

re-write as:
\[y = 1/u^{4} = u^{-4}\]

where:
\[u = 3x + 1\]

so dy/dx = dy/du.du/dx

again, it all then comes down to \[d/dx (x ^{n}) = nx ^{n-1}\]

"chain rule"