How do you differentiate # f(x)= (9-x)(5/x^2 -4) # using the product rule?

Answer 1

The product rule states that for #y=f(x)g(x)#, then #(dy)/(dx)=f'(x)g(x)+f(x)g'(x)#

Remembering the power rule that states #a^-n=1/(a^n)#, we can rewrite it as follows
#f(x)=(9-x)(5x^-2-4)#
#(df(x))=(-1)(5x^-2-4)+(9-x)(-10x^-3)#
#(df(x))/(dx)=(4-5x^-2)+(-90x^-3+10x^-2)#
#(df(x))/(dx)=4+5x^-2-90x^-3#

or, if you prefer,

#(df(x))/(dx)=4+5/x^2-90/x^3#
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Answer 2

To differentiate ( f(x) = (9-x)(\frac{5}{x^2} - 4) ) using the product rule, follow these steps:

  1. Identify the two functions: ( u = 9-x ) and ( v = \frac{5}{x^2} - 4 ).
  2. Find the derivatives of ( u ) and ( v ): ( u' = -1 ) and ( v' = -\frac{10}{x^3} ).
  3. Apply the product rule: ( f'(x) = u'v + uv' ).
  4. Substitute the derivatives and the original functions into the formula: ( f'(x) = (-1)(\frac{5}{x^2} - 4) + (9-x)\left(-\frac{10}{x^3}\right) ).
  5. Simplify the expression: ( f'(x) = -\frac{5}{x^2} + 4 + \frac{10}{x^3} - 9 + x ).
  6. Combine like terms: ( f'(x) = x - 9 -\frac{5}{x^2} + \frac{10}{x^3} + 4 ).
  7. Arrange the terms: ( f'(x) = x - 5/x^2 + 10/x^3 - 5 ).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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