Derivatives of Products and Quotients. Find the derivative?

#y=x(x^2-2)^2#

1 Answer
Jan 10, 2017

#dy/dx=(x^2-2)(5x^2-2), or, 5x^4-12x^2+4#

Explanation:

#y=x(x^2-2)^2#

We first differentiate using the Product Rule.

#dy/dx=xd/dx(x^2-2)^2+(x^2-2)^2d/dx(x)#

#=xd/dx(x^2-2)^2+(x^2-2)^2#

As regards, #d/dx(x^2-2)^2#, we use the Chain Rule.

Let, #u=(x^2-2)^2=t^2, where, t=x^2-2#

So, #u=t^2, t=(x^2-2)#

Thus, #u" is a function of "t, and, t" of "x#

By the Chain Rule, then, #(du)/dx=(du)/dtdt/dx#

Now, #u=t^2 :. (du)/dt=2t, &, t=x^2-2 :. dt/dx=2x#

Utilising these, we have, #d/dx(x^2-2)^2=(du)/dx#

#=(2t)(2x)=4tx=4x(x^2-2)#

Finally, we get,

#dy/dx=4x^2(x^2-2)+(x^2-2)^2, i.e.,#

#dy/dx=(x^2-2){4x^2+(x^2-2)}=(x^2-2)(5x^2-2), or,#

#dy/dx=5x^4-12x^2+4#

Alternatively, without using the Product Rule and Chain Rule, we

can solve the Problem easily, as shown below :

#y=x(x^2-2)^2=x(x^4-4x^2+4)=x^5-4x^3+4x#

#:. dy/dx=5x^4-4(3x^2)+4=5x^4-12x^2+4,# As Before!

Enjoy Maths.!