# Differentiation using the product rule help

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How do I differentiate this?

(x^2-3x+2)(sqrt x)

(x^2-3x+2)(1/2x^-1/2)+(2x-3)(sqrt x)

But what is the next step?

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How do I differentiate this?

(x^2-3x+2)(sqrt x)

(x^2-3x+2)(1/2x^-1/2)+(2x-3)(sqrt x)

But what is the next step?

Expand and simplify

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How do you expand using square roots? I haven't come across that before

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How do you expand using square roots? I haven't come across that before

(x^2-3x+2)(1/2x^-1/2)+(2x-3)(sqrt x)

[(x^4/2)-(3x^2/2)+(2)][(1/2)(x^-1/2)]+[2x^-6/2][x^1/2]

Its a bit hard to show using text, but its just basic exponent multiplication.

Same base = add the exponents. Add the exponents by putting both x^ into the same denominator (in this case 2 on the bottom).

For example: (x^2)((1/2)(x^-1/2) -> (x^4/2)((1/2)(x^-1/2))

it becomes (1/2)(x^[4/2-1/2] => (1/2)(x^3/2)

Its much easier to understand if you write it on paper!

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Its a bit hard to show using text, but its just basic exponent multiplication.

Try using LaTex.

$(x^2-3x+2)({1}/{2}x^{-1/2})+(2x-3)(\sqrt{x})$

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I really still don't understand, mainly due to the set out of the explaination... I apreciate the effort but I can't make sense of it.

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$(x^2-3x+2)(\tfrac{1}{2}x^\frac{-1}{2})+(2x-3)(x^\frac{1}{2})$

$(x^\frac{4}{2}-3x^\frac{2}{2}+2)(\tfrac{1}{2}x^\frac{-1}{2})+(2x^\frac{-6}{2}-3)(x^\frac{1}{2})$

Same base = add the exponents. Add the exponents by putting both x^ into the same denominator (in this case 2 on the bottom).

For example:$(x^2)(\tfrac{1}{2}x^\frac{-1}{2}) -> (x^\frac{4}{2})(\tfrac{1}{2}x^\frac{-1}{2})$

it becomes $(\tfrac{1}{2}x^(\tfrac{4}{2}-\tfrac{1}{2}) => \tfrac{1}{2}x^\frac{3}{2}$

Its much easier to understand if you write it on paper!

Thanks for that link for LaTeX

Lets try this again!

I hope my explaination makes a bit more sense now...

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So can you show me how to get the final answer to this problem? I understand what you mean but I still don't get how to apply it properly

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$(x^2-3x+2)(\tfrac{1}{2}x^\frac{-1}{2})+(2x-3)(x^\frac{1}{2})$

$(x^\frac{4}{2}-3x^\frac{2}{2}+2)(\tfrac{1}{2}x^\frac{-1}{2})+(2x^\frac{2}{2}-3)(x^\frac{1}{2})$

Multiply the terms through

$(\tfrac{1}{2}x^\frac{3}{2}-\tfrac{3}{2}x^\frac{1}{2}+\tfrac{2}{2}x^\frac{-1}{2})+(2x^\frac{3}{2}-3x^\frac{1}{2})$

Group terms with the same exponent ie:$x^\frac{1}{2}$

$(\tfrac{1}{2}x^\frac{3}{2}+2x^\frac{3}{2})+(-\tfrac{3}{2}x^\frac{1}{2}-3x^\frac{1}{2})+x^\frac{-1}{2}$

Only have one of each 'type' of x

$(\tfrac{1}{2}x^\frac{3}{2}+\tfrac {4}{2}x^\frac{3}{2})+(-\tfrac{3}{2}x^\frac{1}{2}-\tfrac{6}{2}x^\frac{1}{2})+x^\frac{-1}{2}$

$(\tfrac{5}{2}x^\frac{3}{2}-\tfrac{9}{2}x^\frac{1}{2}+x^\frac{-1}{2})$

Reduced form:

$\tfrac{1}{2}x^\frac{-1}{2}(5x^\frac{4}{2}-9x^\frac{2}{2}+2)$

$\tfrac{1}{2}x^\frac{-1}{2}(5x^2-9x+2)$ (done here)

If that ended up to be -2 instead:

$\tfrac{1}{2}x^\frac{-1}{2}(5x^2-9x-2)$

It can be factored like this:

$\tfrac{1}{2}x^\frac{-1}{2}(5x+1)(x-2)$

I think that's right.

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Yeah, I got the same. Although I'd say it'd be easier to expand out before differentiating, and avoiding the chain rule altogether.

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