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g(x)=(x^3-8)*x^2+1/x^2-1

Differentiate the function
 

Comments

Gisselle,
 
It would help if you cleared up how the order of operations applies to this problem.
 
For example, do you mean to multiply (x^3-8) times (x^2) or (x^2+1)?
 
Or how about the last terms, do you want to have 1 divided by (x^2) or (x^2+1)
 
As currently constructed your function is
[(x^3-8)(x^2)] + [1/(x^2)] - [1]
 
I doubt this is your actual function.  But if it is you can use the addition rule and break it into three derivatives
[(x^3-8)(x^2)]
You need to you the product rule v du + u dv where v is x^2 and u is x^3-8
 
[1/(x^2)]
You can use the power rule for this using the function x^(-2)
 
[1]
Derivative of a constant is 0
 
Why do you believe she wrote it incorrectly and doubt that
d/dx [ -1 + 1/x2 + x2 (-8+x3) ]  is what she meant?
 
I assumed in my answer that she wrote the problem down correctly and I applied the order of operations to the equation the way she wrote it. I hope a calculus student knows order of operations.
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1 Answer

d/d(x) [ (x3-8) • x2 + 1/x2 - 1 )
rewrite
d/dx [ -1 + 1/x2 + x2 (-8+x3) ]
 
differentiate the sum term by term
d/dx (-1) + d/dx (1/x2) + d/dx [x2 (-8+x3) ]
 
derivative of -1 = 0
power rule
d/dx(xn) = n•xn-1
n= -2,  d/dx (1/x2) = d/dx (x-2) = -2x3
d/dx = (x2 (-8+x3)) + (-2/x3)
 
product rule
d/dx (uv)= v du/dx + u dv/dx
u = x2,  v = x3-8
 = (-2/x3) + (x3 - 8) d/dx(x2) + x2 d/dx(-8+x3)
 
power rule
d/dx(xn) = n•xn-1
n= -2,  d/dx (x2) = 2x
= (-2/x3) + x2 (d/dx(-8+x3)) + (-8+x3)(2x)
 
differentiate the sum term by term
= (-2/x3) + (2x)(-8+x3) + (d/dx(-8) + d/dx(x3) x2
 
derivative of -8 is 0
Product rule
= (-2/x3) + (2x)(-8+x3) + (3x2) • x2
 
Simplify
= (-2/x3) + (2x)(-8+x3) + 3x4
= (-2/x3) + (-16x) + (2x4) + 3x4
= 5 x4 - 2 / x3 - 16 x

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