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

Question

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

Samantha Adkison · Accepted Answer

#f'(x)=5*(x^2-18x+4)/((x-2)^2*(x+2)^2)# After the product rule #(uv)'=u'v+uv'# we get #f'(x)=-5/(x^2-4)+(9-x)*(-5)*(x^2-4)^(-2)*2*x# which simplifies to #f'(x)=5*(x^2-18x+4)/((x-2)^2*(x+2)^2)#

Evelyn Agard · Answer

#f'(x)=(5(x^2 -18x+4))/(x^2 -4)^2 #

#f(x)= (9-x)(5/(x^2 -4))#

Apply product rule,

#f'(x)= [d/dx(9-x)] (5/(x^2 -4))+[d/dx(5/(x^2 -4))] (9-x)#

Differentiate inner terms,

#f'(x)=(-1) (5/(x^2 -4))+(-(10x)/(x^2 -4)^2) (9-x)#

Simplify,

#f'(x)=-5/(x^2 -4)-(10x(9-x))/(x^2 -4)^2 #

Common denominator,

#f'(x)=-(5(x^2 -4))/(x^2 -4)^2-(10x(9-x))/(x^2 -4)^2 #

Simplify,

#f'(x)=-(5x^2 -20)/(x^2 -4)^2-(90x-10x^2)/(x^2 -4)^2 #

Combine,

#f'(x)=(-5x^2 +20-90x+10x^2)/(x^2 -4)^2 #

Simplify,

#f'(x)=(5x^2 -90x+20)/(x^2 -4)^2 #

Factorise,

#f'(x)=(5(x^2 -18x+4))/(x^2 -4)^2 #

Charles Arocha · Answer

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

1. Identify the two functions being multiplied: $ u(x) = 9 - x $ and $ v(x) = \frac{5}{{x^2 - 4}} $.
2. Compute the derivatives of $ u(x) $ and $ v(x) $: $ u'(x) = -1 $ and $ v'(x) = \frac{{-10x}}{{(x^2 - 4)^2}} $.
3. Apply the product rule: $ f'(x) = u'(x)v(x) + u(x)v'(x) $.
4. Substitute the values into the product rule formula: $ f'(x) = (-1)\left(\frac{5}{{x^2 - 4}}\right) + (9 - x)\left(\frac{{-10x}}{{(x^2 - 4)^2}}\right) $.
5. Simplify the expression to get the derivative of $ f(x) $: $ f'(x) = -\frac{5}{{x^2 - 4}} - \frac{{(9 - x)(10x)}}{{(x^2 - 4)^2}} $.