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I'm having trouble simplifying these questions, particularly when they involve square roots of $x$.

Differentiate the following with respect to $x$ and simplify:

$y=(x+2)x^\frac{3}{2}$

My attempt:

Using product rule: $u=x^\frac{3}{2}, v=(x+2)$ therefore $\frac{du}{dx}=\frac{3}{2}x^\frac{1}{2}, \frac{dv}{dx}=1\\$

$\frac{dy}{dx}= \frac{3\sqrt{x}}{2}(x+2)+(\sqrt{x})^3$

Factorise:$\sqrt{x}[\frac{3}{2}(x+2)+(\sqrt{x})^2]\\\sqrt{x}[\frac{3}{2}x+3+x]$

The given answer is $\frac{\sqrt{x}}{2}(5x+6)$, which I can't achieve and I can't understand why the denominator of 2 is a common factor.

Paddington
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    They pulled $\frac{1}{2}$ from each term in the brackets. Your answer is correct – pwerth Apr 05 '18 at 19:17
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    Another, maybe simpler way to calculate this derivative is to expand your function. $(x+2)x^{3/2} dx = x^{5/2} + 2 x^{3/2} dx = {5\over 2} x^{3/2} + 3 x^{1/2} $ – Fibonacci Apr 05 '18 at 19:21
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    Paddington. Correct. √x((3/2)x +(2/2)x +3)=√x((5/2)x+3)= √x((5/2)x+6/2)=(√x/2)(5x+6). – Peter Szilas Apr 05 '18 at 19:28

2 Answers2

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You derivation is correct indeed note that

$$\sqrt{x}\left(\frac{3}{2}x+3+x\right)=\frac{\sqrt{x}}{2}\left(3x+6+2x\right)=\frac{\sqrt{x}}{2}\left(5x+6\right)$$

user
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  • Thanks.I'm also confused why the second and third $\frac{\sqrt{x}}{2}$ isn't $2\sqrt{x}$. To factorise surely you're either dividing by $\frac{1}{2}$ or multiplying by 2? – Paddington Apr 06 '18 at 10:43
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    Let consider $\sqrt{x}\left(\frac{3}{2}x+3+x\right)=\sqrt{x}\left(\frac{3x+6+2x}{2}\right)$ and then factor out $\frac12$. – user Apr 06 '18 at 10:58
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A general rule: first simplify, then differentiate. The function in question is a product, namely $$y=(x+2)x^\frac{3}{2}.$$ First simplify it to a sum $$y=x^{\frac{5}{2}}+2x^\frac{3}{2}$$ and then differentiate $$y’=\frac{5}{2}x^{\frac{3}{2}}+3x^{\frac12}.$$

Michael Hoppe
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    Would it be possible to simplify first if the bracket was raised to a fractional exponent? – Paddington Apr 06 '18 at 12:07
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    In one word: No. – Michael Hoppe Apr 07 '18 at 06:24
  • Well, if you're very careful, then $y = (ax+b)^{p/q}x^k$ can be "simplified" to $y=((ax+b)(x^k)^{q/p})^{p/q} = (ax^{(p+kq)/p}+bx^{kq/p})^{p/q}$, so that you no longer need the product rule. But when dealing with fractional exponents make sure you're not introducing or removing assumptions artificially, such as the requirement $A \geq 0$ in order for $A^{1/n}$ to make sense. – Stef Jun 22 '23 at 11:22