can anyone differentiate this for me using the quotient rule?

Question

can anyone differentiate this for me using the quotient rule: f(x)=2x^5 +x^2/ x. The answer is 8x+1. I get 32x^3-24x^2+2/ (2x-1)^2. Please show steps thanks!

Answer 1

I'll assume that it is (2x^5 + x^2) / x. You could easly simplify it by taking an x out of the numerator and canceling it out with the one from the denominator, and if it is 2x^5 + (x^2 / x), then you could just cancel one x from the numerator of the fraction with the denominator. But, i'll assume it is the first one, and do this with the quotient rule.
the quotient rule is "low D high - high D low over low low" (well, thats how i remember it)

Original equation:
f(x) = (2x^5 + x^2) / x

The derivative:
f'(x) = ( (x) D (2x^5 + x^2) - (2x^5 + x^2) D (x) ) / ( (x) (x) )

The derivative of 2x^5 + x^2, using the power rule, is 2*5x^(5-1) + 2*x^(2-1) = 10x^4 + 2x, and the derivative of x is 1.
f'(x) = ( (x) (10x^4 + 2x) - (2x^5 + x^2) * 1 ) / ( x^2 )

Now i'll expand the stuff
f'(x) = ( 10x^5 + 2x^2 - 2x^5 - x^2 ) / ( x^2 )

Combine like terms
f'(x) = ( 8x^5 + x^2 ) / ( x^2 )

Now, factoring out an x^2 from the numerator and canceling it with the denominator:
f'(x) = 8x^3 + 1

And theres the derivative with the quotient rule.

Answer 2

2x^5+x²/x = 2x^5+x, so the derivative is 10x^4+1. Even if we assume you meant to write (2x^5+x²)/x, this still reduces to 2x^4+x, so the derivative would be 8x³+1, not 8x+1. I'm going to assume you meant the latter, since then the correct answer would bear more similarity to what you actually wrote.

Under the quotient rule, (f/g)' = (f'g-g'f)/g². In this case, f=(2x^5+x^2) and g=x. So f'=10x^4+2x, and g'=1. So using the quotient rule:

d((2x^5+x²)/x)/dx = ((10x^4+2x)*x-1*(2x^5+x^2))/x²

Simplifying:

((10x^4+2x)*x-1*(2x^5+x²))/x²
(10x^5+2x²-2x^5-x²))/x²
8x^5+x²)/x²
8x³+1

Which is the same answer we got by simplifying first.

Answer 3

The answer should be 8x^3 + 1
It's because f(x) = 2x^4 + x
f'(x) = 8x^3 + 1

But if you really wish to use the quotient rule.
u = 2x^5 +x^2
v = x
du/dx = 10x^4 + 2x
dv/dx = 1

f' = (v*du/dx - u*dv/dx) / v^2
= 1/x^2 * [ 10x^5 + 2x^2 - 2x^5 - x^2]
= 1/x^2 * [8x^5 + x^2]
= 8x^3 + 1

Answer 4

the rule is [ g(x)f'(x) - f(x)g'(x) ] / [g(x)]² or (vu'-uv')/v²
so in u/v, u = 2x^5 +x^2, u'= 10x^4 + 2x
v=x, v'=1
so we have [x(10x^4 + 2x) - (2x^5 +x^2)(1)]/[x^2]
[10x^5 + 2x^2 - 2x^5 - x^2]/[x^2]
(8x^5+x^2)/(x^2)
= 8x^3 + 1 not 8x+1

Answer 5

f= 2x^5+x^2 f'= 10x^4 +2x
g= x g' =1
quotient rule = (gf' -fg')/g^2
x(10x^4 +2x) - 2x^5+x^2
8x^5 - x^2/x^2
8x^3 +1

Answer 6

First of all, you would wish to preclude utilising quotient/product laws for the reason that the ones make matters extra tricky. one million/x^n = (x^-n), so you'll rewrite the query as (two/three) * x^-five and differentiate that utilising the energy rule: (cx^n)' = cnx^(n-one million)

Answer 7

f(x)=(2x^5 +x^2)/ x.
f ' (x) =[ (x)(10x^4 + 2x) - (2x^5 +x^2)(1)] /x^2
f ' (x) =[ 10x^5 + 2x^2 - 2x^5 -x^2]/x^2
f ' (x) =[ 8x^5 + x^2]/x^2
f ' (x) =[x^2 ( 8x^3 +1)]/x^2
f ' (x) =8x^3 +1

QED..check ans again.