If G Is Twice Differentiable Function And F(x)=xg(x^2), How Do You Find F ...

If g is twice differentiable function and #f(x)=xg(x^2)#, how do you find f'' in terms of g, g', and g''? Calculus Basic Differentiation Rules Chain Rule

1 Answer

Steve M Nov 3, 2016

# f''(x) = 4x^3g''(x^2) + 6xg'(x^2)#

Explanation:

First we need to use the product rule, as follows:

Differentiating Once: # f(x)=xg(x^2) # # :. f'(x)=(x)(d/dxg(x^2)) + (d/dxx)(g(x^2))# # :. f'(x)=x(d/dxg(x^2)) + (1)(g(x^2))# # :. f'(x)=x(d/dxg(x^2)) + g(x^2)#

Then By the chain rule, we have: # d/dxg(x^2) = g'(x^2)(d/dxx^2) # # d/dxg(x^2) = g'(x^2)(2x) # # :. d/dxg(x^2) = 2xg'(x^2) # .... [1]

Substituting [1] gives us: # :. f'(x)=x(2xg'(x^2)) + g(x^2)# # :. f'(x)=2x^2g'(x^2) + g(x^2)#

Differentiating a second time, (again using the product rule): # f''(x) = (2x^2)(d/dxg'(x^2)) + (d/dx2x^2)(g'(x^2)) + (d/dxg(x^2))#

Again, Substituting [1] gives us: # f''(x) = (2x^2)(d/dxg'(x^2)) + (4x)(g'(x^2)) + (2xg'(x^2))# # f''(x) = 2x^2(d/dxg'(x^2)) + 6xg'(x^2)#

And, again by the chain rule, we have: # d/dxg'(x^2) = (g''(x^2))(d/dxx^2) # # :. d/dxg'(x^2) = (g''(x^2))(2x) # # :. d/dxg'(x^2) = 2xg''(x^2) # .... [2]

Substituting [2] gives us: # f''(x) = 2x^2(2xg''(x^2)) + 6xg'(x^2)# # f''(x) = 4x^3g''(x^2) + 6xg'(x^2)#

Answer link

Related questions

• What is the Chain Rule for derivatives?
• How do you find the derivative of #y= 6cos(x^2)# ?
• How do you find the derivative of #y=6 cos(x^3+3)# ?
• How do you find the derivative of #y=e^(x^2)# ?
• How do you find the derivative of #y=ln(sin(x))# ?
• How do you find the derivative of #y=ln(e^x+3)# ?
• How do you find the derivative of #y=tan(5x)# ?
• How do you find the derivative of #y= (4x-x^2)^10# ?
• How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ?
• How do you find the derivative of #y= ((1+x)/(1-x))^3# ?

See all questions in Chain Rule

Impact of this question

25468 views around the world You can reuse this answer Creative Commons License