Basic quotient rule problem but root is confusing....

[EJnr]

Good Day,

I am working on some basic calculus problems currently and am unable to solve the following with the quotient rule as required (it seems that the introduction of the root is confusing me).

Y = X^3/square root(X-1)

A simple problem Im sure but I cannot seem to reconcile the roots and reach the advised solution of

X^2(5X-6)/2(X-1)^3/2

If anyone would be able to advise on the solution line by line it would be most appreciated - I am sure I am just missing out on some common rules (besides the quotient rule itself!) and simplifying that will tidy up the problem to arrive at the solution?

Kind Regards and Sincere Thanks in advance,

EJnr

 

[Deleted member 4993]

Good Day,

I am working on some basic calculus problems currently and am unable to solve the following with the quotient rule as required (it seems that the introduction of the root is confusing me).

Y = X^3/square root(X-1)

A simple problem Im sure but I cannot seem to reconcile the roots and reach the advised solution of

X^2(5X-6)/2(X-1)^3/2

If anyone would be able to advise on the solution line by line it would be most appreciated - I am sure I am just missing out on some common rules (besides the quotient rule itself!) and simplifying that will tidy up the problem to arrive at the solution?

Kind Regards and Sincere Thanks in advance,

EJnr

I think you are looking for the derivative of the given function.

\(\displaystyle f(x) \ = \ \frac{x^3}{\sqrt{x-1}}\)

Quotient rule (which I see as a corollary of the product rule):

\(\displaystyle f(x) \ = \ \frac{g(x)}{h(x)}\)

\(\displaystyle \frac{df(x)}{dx} \ = \ \frac{g'(x)\cdot h(x) \ - \ h'(x)\cdot g(x)}{[h(x)]^2}\)........................(1)

In your case:

\(\displaystyle g(x) \ = \ x^3\).............(2)...............\(\displaystyle g'(x) \ = \ 3x^2\).................................(3)

\(\displaystyle h(x) \ = \ \sqrt{x-1}\).......(4)............\(\displaystyle h'(x) \ = \ \frac{1}{2\sqrt{x-1}}\)................(5)

Now put (2), (3), (4) and (5) into (1) and simplify.

If you are still stuck - please show work indicating exactly where you are stuck.

 

[soroban]

Hello, EJnr!

\(\displaystyle \text{Differentiate: }\:Y \:=\: \frac{x^3}{\sqrt{x-1}}\)

\(\displaystyle \text{Answer: }\:y' \;=\;\frac{x^2(5x-6)}{2(x-1)^{\frac{3}{2}}}\)

\(\displaystyle \text{We have: }\;y \;=\;\frac{x^3}{(x-1)^{\frac{1}{2}}}\)

\(\displaystyle \text{Quotient Rule: }\;y' \;=\;\frac{(x-1)^{\frac{1}{2}}\!\cdot\!3x^2 \:-\: x^3\!\cdot\!\frac{1}{2}(x-1)^{\text{-}\frac{1}{2}}}{x-1}\)

\(\displaystyle \text{Factor: }\quad y' \;=\;\frac{\frac{1}{2}(x-1)^{\text{-}\frac{1}{2}}\bigg[6x^2(x-1) - x^3\bigg]}{x-1} \;=\;\frac{1}{2}\cdot\frac{1}{(x-1)^{\frac{1}{2}}}\cdot\frac{6x^3 - 6x^2 - x^3}{x-1}\)

. . . . . . . . . \(\displaystyle =\;\frac{5x^3-6x^2}{2(x-1)^{\frac{3}{2}}} \;=\;\frac{x^2(5x-6)}{2(x-1)^{\frac{3}{2}}}\)

 

[EJnr]

Thank you so much Subhotosh and Soroban for your prompt responses. Yes that all makes sense now.

Thank you also Soroban for perhaps sensing that it was the simplification and specifically the factoring that I was unable to do. The rules themselves are of course straight forward but like many I do find the simplification and factorisation can trip me up - can anyone recommend a good website or source for practicing and mastering this process (of simplification and factorising) at this level and beyond? Also if you could advise how to enter the maths script as you both did that would be much appreciated also.

Once again many thanks for your assistance.

Kind Regards,

EJnr