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