# Can anyone explain this process of solving? (Differentiation)

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?

• apply quotient and chain rule together Aug 15 at 10:12
• They simply factor out $x(x+1)^2$ from both terms. Think of it as a reverse of the distribution law Aug 15 at 10:25
• @Cathedral oh so is that why (x+1)^3 is now (x+1)?
– Hal
Aug 15 at 10:28
• @TobyMak thanks for the answer! It seems easier now that a lot has explained
– Hal
Aug 15 at 10:51
• @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. Aug 15 at 18:57

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:

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

• 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. Aug 15 at 18:49

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

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: $$y'(x)=\frac{x(x+1)^2[3x-2(x+1)]}{x^4}=\frac{(x+1)^2(x-2)}{x^3}$$

• @Teepeemm Thank you, i edited this answer. Aug 16 at 2:56