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