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

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Answer 1 · 2018-05-09T05:10:10

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

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

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

#(dy)/(dx)+2xy^3+x^2*3y^2(dy)/(dx)=0#

or #(dy)/(dx)[1+3x^2y^2]=-2xy^3#

or #(dy)/(dx)=-(2xy^3)/(1+3x^2y^2)#

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

Answer 2 · 2018-05-09T05:34:49

The answer is #=-(2x(f(x))^3)/(1+3x^2(f(x))^2)#

The function is

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

Let #y=f(x)#

Then,

#y+x^2y^3-30=0#

Let

#f(x,y)=y+x^2y^3-30#

Then,

#dy/dx=-((delf)/(delx))/((delf)/(dely))#

#(delf)/(delx)=2xy^3#

#(delf)/(delx)=1+3y^2x^2#

#dy/dx=-(2xy^3)/(1+3x^2y^2)=-(2x(f(x))^3)/(1+3x^2(f(x))^2)#