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Question 1010460: differentiate the following function:
f(x)=(x^2+1/x^3 )(x^3-x^2+1)
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! f(x)=(x^2+1/x^3 )(x^3-x^2+1)
The derivative of f(x) is f'(x)
f'(x) = d/dx((1/x^3+x^2) (1-x^2+x^3))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = 1/x^3+x^2 and v = x^3-x^2+1
f'(x) = (x^3-x^2+1) d/dx(1/x^3+x^2)+(x^2+1/x^3) d/dx(1-x^2+x^3)
Differentiate the sum term by term
f'(x) = (1/x^3+x^2) (d/dx(1-x^2+x^3))+(1-x^2+x^3) d/dx(1/x^3)+d/dx(x^2)
Use the power rule, d/dx(x^n) = n x^(n-1), where n = -3: d/dx(1/x^3) = d/dx(x^(-3)) = -3 x^(-4)
f'(x) = (1/x^3+x^2) (d/dx(1-x^2+x^3))+(1-x^2+x^3) (d/dx(x^2)+(-3)/x^4)
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x
f'(x) = (1/x^3+x^2) (d/dx(1-x^2+x^3))+(1-x^2+x^3) (-3/x^4+2 x)
Differentiate the sum term by term and factor out constants
f'(x) = (-3/x^4+2 x) (1-x^2+x^3)+(1/x^3+x^2) d/dx(1)-d/dx(x^2)+d/dx(x^3)
The derivative of 1 is zero
f'(x) = (-3/x^4+2 x) (1-x^2+x^3)+(1/x^3+x^2) (-(d/dx(x^2))+d/dx(x^3)+0)
Simplify the expression
f'(x) = (-3/x^4+2 x) (1-x^2+x^3)+(1/x^3+x^2) (-(d/dx(x^2))+d/dx(x^3))
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x
f'(x) = (-3/x^4+2 x) (1-x^2+x^3)+(1/x^3+x^2) (d/dx(x^3)-2 x)
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 3: d/dx(x^3) = 3 x^2
f'(x) = (-3/x^4+2 x) (1-x^2+x^3)+(1/x^3+x^2) (-2 x+3 x^2)
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f'(x) = (1/x^3+x^2) (-2 x+3 x^2)+(-3/x^4+2 x) (1-x^2+x^3)
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