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I'm at differentiation of algebraic functions. There's an example in the module that I couldn't quite get how it led to that.

$y=\frac {(x+1)^3}{x^2}$

It's solved by using a combination of quotient and power rules. I'll enumerate how it's solved.

$(1) y′= \frac{(x^2(3(x+1)^2))–(x+1)^32x}{x^4}$

$(2) y′= \frac{x(x+1)^2 (3x–2(x+1))}{x^4}$

$(3) y′= \frac {(x+1)^2 (3x-2x-2)}{x^3}$

$(4) y′= \frac {(x+1)^2 (x-2)}{x^3}$

How did $(2)$ came to be? Why was it done like that? I just can't wrap my head around it. It just looked like it skipped a couple steps (at least to me). Can anyone help me with this?

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  • $\begingroup$ apply quotient and chain rule together $\endgroup$
    – abcdefu
    Aug 15 at 10:12
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    $\begingroup$ They simply factor out $x(x+1)^2$ from both terms. Think of it as a reverse of the distribution law $\endgroup$
    – Cathedral
    Aug 15 at 10:25
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    $\begingroup$ @Cathedral oh so is that why (x+1)^3 is now (x+1)? $\endgroup$
    – Hal
    Aug 15 at 10:28
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    $\begingroup$ @TobyMak thanks for the answer! It seems easier now that a lot has explained $\endgroup$
    – Hal
    Aug 15 at 10:51
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    $\begingroup$ @Cathedral My guess is that the "next step" Hal was expecting after (1) would be to completely expand the numerator. Which ends up with a cubic that needs factoring, so it's a more complicated approach. @ Hal: a takeaway lesson is that sometimes the factored form of an expression is better, and sometimes the expanded form is better. Life will be a bit easier if you take a step back and think about which is better for each particular instance. $\endgroup$
    – Teepeemm
    Aug 15 at 18:57

3 Answers 3

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If it gets too messy, you can always use logarithmic differentiation. Take the natural logarithm of both sides to get:

$$\ln y = \ln((x+1)^3) - \ln(x^2)$$ $$\ln y = 3 \ln(x+1) - 2 \ln x$$

and so using the chain rule, we have that $\frac{d}{dx} (\ln y)$, where $y$ is a function of $x$), is $\frac{1}{y} \cdot y' = \frac{y'}{y}$:

$$\frac{y'}{y} = \frac{3}{x+1} - \frac{2}{x} = \frac{3x - 2(x+1)}{x(x+1)} = \frac{x-2}{x(x+1)}$$

so using what $y$ is, we have that:

$$y' = \frac{dy}{dx} = \frac{x-2}{x(x+1)} \frac{(x+1)^3}{x^2} = \frac{(x-2)(x+1)^2}{x^3}$$

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    $\begingroup$ If OP is just learning how to differentiate algebraic functions, I don't know that they've covered differentiating the natural logarithm and the chain rule, let alone logarithmic differentiation. $\endgroup$
    – Teepeemm
    Aug 15 at 18:49
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Here is one way to solve it. Use the fact that the derivative of $(ax+b)^n=n(ax+b)^{n-1}(ax+b)'$. This just follows from the chain rule.

Rewrite the equation as $y=(x+1)^3x^{-2}$. Then apply the product rule. The product rule says $(f(x)g(x))'=f(x)g'(x)+g(x)f'(x)$.

Thus, we get $((x+1)^3x^{-2})'=((x+1)^3\cdot-2x^{-3})+3(x+1)^2x^{-2}$.

Writing this out in a clean way we get: $((x+1)^3x^{-2})'=\frac {-2(x+1)^3} {x^3}+\frac {3(x+1)^2} {x^2}=\frac {-2(x+1)^3+3x(x+1)^2} {x^3}$.

Now factor out $(x+1)^2$ to get $\frac {(x+1)^2(-2(x+1)+3x)} {x^3}=\frac {(x+1)^2(-2x-2+3x)} {x^3}=\frac {(x+1)^2(x-2)} {x^3}$

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Before we start, let's recall the first derivative of a quotient:

Let: $$y(x)=\frac{u(x)}{v(x)}, v(x)\ne 0$$ then:

$$y'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v^2(x)}$$

Apply for your case:

$$y(x)=\frac{(x+1)^3}{x^2}$$ then: $$y'(x)=\frac{[(x+1)^3]'x^2-(x+1)^3[x^2]'}{x^4}=\frac{3(x+1)^2x^2-2x(x+1)^3}{x^4}$$ now use the common factor $x(x+1)^2$ then: $$y'(x)=\frac{x(x+1)^2[3x-2(x+1)]}{x^4}=\frac{(x+1)^2(x-2)}{x^3}$$

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  • $\begingroup$ @Teepeemm Thank you, i edited this answer. $\endgroup$
    – OnTheWay
    Aug 16 at 2:56

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