How do you differentiate f(x) + x^2 [f(x)]^3 = 30f(x)+x2[f(x)]3=30?

2 Answers
Shwetank Mauria
May 9, 2018

(df(x))/(dx)=-(2x[f(x)]^3)/(1+3x^2[f(x)]^2)df(x)dx=−2x[f(x)]31+3x2[f(x)]2

Explanation:

Let y=f(x)y=f(x), then we can write f(x)+x^2[f(x)]^3=30f(x)+x2[f(x)]3=30 as

y+x^2y^3=30y+x2y3=30 and differentiating it implicitly

(dy)/(dx)+2xy^3+x^2*3y^2(dy)/(dx)=0dydx+2xy3+x2⋅3y2dydx=0

or (dy)/(dx)[1+3x^2y^2]=-2xy^3dydx[1+3x2y2]=−2xy3

or (dy)/(dx)=-(2xy^3)/(1+3x^2y^2)dydx=−2xy31+3x2y2

or (df(x))/(dx)=-(2x[f(x)]^3)/(1+3x^2[f(x)]^2)df(x)dx=−2x[f(x)]31+3x2[f(x)]2

Narad T.
May 9, 2018

The answer is =-(2x(f(x))^3)/(1+3x^2(f(x))^2)=−2x(f(x))31+3x2(f(x))2

Explanation:

Another method for implicit differentiation

The function is

f(x)+x^2(f(x))^3=30f(x)+x2(f(x))3=30

Let y=f(x)y=f(x)

Then,

y+x^2y^3-30=0y+x2y3−30=0

Let

f(x,y)=y+x^2y^3-30f(x,y)=y+x2y3−30

Then,

dy/dx=-((delf)/(delx))/((delf)/(dely))dydx=−∂f∂x∂f∂y

(delf)/(delx)=2xy^3∂f∂x=2xy3

(delf)/(delx)=1+3y^2x^2∂f∂x=1+3y2x2

dy/dx=-(2xy^3)/(1+3x^2y^2)=-(2x(f(x))^3)/(1+3x^2(f(x))^2)dydx=−2xy31+3x2y2=−2x(f(x))31+3x2(f(x))2

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