How to differentiate $y=(x+1)^3/x^{3/2}$ and $y=2x^4/(b^2-x^2)$
I need to solve a list of derivatives to help me on an exam; however, I'm in doubt when they use another variable (constant) or when I have a fraction with functions that use the power rule.
For example:
$$y = \frac{(x + 1)^3}{x^{3/2}}.$$
This one I tried having $u = x + 1$, deriving it and having $v = x^{3/2}$, but I couldn't get the right answer.
And:
$$y = \frac{2x^4}{b^2 - x^2}.$$
And this one I tried to separate the constant, like $y = \frac{1}{b^2} \frac{dx}{dy} \frac{2x^4}{-x^2}$, but also coudn't find the answer.
Can you give a tip on how to find which rule to apply to these kind of derivatives?
$\endgroup$ 22 Answers
$\begingroup$You really only need the quotient rule and the power rule.
For the quotient rule, when $y=\frac{f(x)}{g(x)}$ (where $g(x)\neq 0$), we have $y'=\frac{f'g-fg'}{g^2}$.
Problem 1: Differentiate $y = \frac{(x + 1)^3}{x^{3/2}}$ with respect to $x$.
Solution. We have $y=\frac{f(x)}{g(x)}$ where $f(x)=(x+1)^3$ and $g(x)=x^{3/2}$. Thus, we also have $f'(x)=3(x+1)^2$ and $g'(x)=\frac{3}{2}x^{1/2}$. Using the quotient rule, we see that $$ y' = \frac{3(x+1)^2\cdot x^{3/2}-(x+1)^3\cdot\frac{3}{2}x^{1/2}}{(x^{3/2})^2}. $$
Problem 2: Differentiate $y = \frac{2x^4}{b^2 - x^2}$ with respect to $x$.
Solution. We have $y=\frac{f(x)}{g(x)}$ where $f(x)=2x^4$ and $g(x)=b^2-x^2$. Thus, we also have $f'(x)=8x^3$ and $g'(x)=-2x$. Using the quotient rule, we see that $$ y'=\frac{8x^3\cdot(b^2-x^2)-2x^4\cdot(-2x)}{(b^2-x^2)^2}. $$
If required, all that is left to do is for you to simplify.
$\endgroup$ 5 $\begingroup$When the expression contains products and ratioé, logarithmic differentiation make life much easier.
Let us consider the first problem $$y = \frac{(x + 1)^3}{x^{3/2}}$$ Taking logarithms $$\log(y)=3\log(x+1)-\frac32\log(x)$$ Now, differentiate both sides $$\frac{y'}y=\frac3{x+1}-\frac3{2x}=\frac{3 (x-1)}{2 x (x+1)}$$ $$y'=\frac{3 (x-1)}{2 x (x+1)}y=\frac{3 (x-1)}{2 x (x+1)} \frac{(x + 1)^3}{x^{3/2}}=\frac{3 (x-1) (x+1)^2}{2 x^{5/2}}$$
Do the same for the second problem $$y = \frac{2x^4}{b^2 - x^2}$$ $$\log(y)=\log(2)+4\log(x)-\log(b^2-x^2)$$ $$\frac{y'}y=0+\frac 4x+\frac{2x}{b^2-x^2}=\frac{2(2b^2-x^2)}{x(b^2-x^2)}$$ $$y'=\frac{2(2b^2-x^2)}{x(b^2-x^2)}y=\frac{2(2b^2-x^2)}{x(b^2-x^2)}\frac{2x^4}{b^2 - x^2}=\frac{4(2b^2-x^2)x^3}{(b^2-x^2)^2}$$
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