# If y=(1+x^2)^3 find dy/dx.

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### 3 Answers

We have y=(1+x^2)^3

We have to find dy/dx. We can use the chain rule here.

dy/dx = 3(1 + x^2)^2*2x

=> dy/dx = 6x(1 + x^2)^2

The required result is

**6x*(1 + x^2)^2**

We'll apply chain rule to evaluate dy/dx.

dy/dx = (dy/dt)*(dt/dx)

We'll put 1+x^2 = t

y = t^3

We'll differentiate with respect to t:

dy/dt = d(t^3)/dt

dy/dt = 3t^2

dt/dx = d(1+x^2)/dx

We'll differentiate with respect to t:

dt/dx = 2x

dy/dx = 3t^2*2x = 6x*t^2

We'll substitute back t:

**dy/dx = 6x(1 + x^2)^2**

y = (1+x^2)^3.

We have to find dy/dx

We use d/dx {u(v(x))} = (du/dv) (dv/dx)

Let v(x) = 1+x^2

d/dx {v(x)} = d/ dx {1+x^2} = d/dx(1)+ d/ dx(x^2) = = 0+2x

Therefore d/dx{1+x^2)^3 = 3(1+x^2^(3-1)*d/dx (1+x^2)

d/dx(1+x^2)^3 = 3(1+x^2)^2*2x

**Therefore dy/dx = d/dx(1+x^2)^3 = 6x(1+x^2)^2.**