D/dx x(x^2 +1) ^{1 /2}/(x+1) ^{2 /3}

[karush]

​$\large{242.q2.1c}$
Find the derivative
$$\displaystyle
y=\frac{x(x^2 +1) ^{1/2}}{(x+1)^{2/3}}$$
I know this is using the quotient rule but can't seem simplify this first

 

[kaliprasad]

​$\large{242.q2.1c}$
Find the derivative
$$\displaystyle
y=\frac{x(x^2 +1) ^{1/2}}{(x+1)^{2/3}}$$
I know this is using the quotient rule but can't seem simplify this first

I do not see anything in common . So this cannot be simplified fisrt

you can take it of the form $\frac{u}{v}$ and then u as a product.

 

[soroban]

I suggest Logarithmic Differentiation.

 

[karush]

I suggest Logarithmic Differentiation.

I think that's what was intended but missed the lecture on it

 

[MarkFL]

I think that's what was intended but missed the lecture on it

Take the natural log of both sides, apply the rules of logs to the right side, then implicitly differentiate. :)

 

[soroban]

$$\text{Find the derivative :}$$
$$\displaystyle
y=\frac{x(x^2 +1) ^{1/2}}{(x+1)^{2/3}}$$

$$\text{Take logs: }\; \ln y \;=\;\ln\left[\frac{x(x^2+1)^{\frac{1}{2}}}{(x+1)^{\frac{2}{3}}}\right] \;=\;\ln x + \ln(x^2+1)^{\frac{1}{2}} - \ln(x+1)^{\frac{2}{3}}$$

. .. .. .. . . . $$\ln y \;=\;\ln x + \tfrac{1}{2}\ln(x^2+1) - \tfrac{2}{3}\ln(x+1)$$

. . $$\frac{y'}{y} \;=\;\frac{1}{x} + \frac{1}{2}\frac{2x}{x^2+1} = \frac{2}{3}\frac{1}{x+1}$$

. . . . $$\text{ . . . etc. . . . }$$

 

[karush]

$\displaystyle
\ln y \;=
\;\ln\left[\frac{x(x^2+1)^{\frac{1}{2}}}{(x+1)^{\frac{2}{3}}}\right] \\
=\ln x
+ \frac{1}{2}\ln(x^2+1) - \frac{2}{3}\ln(x+1)$

$\displaystyle
y' =y \left[\frac{1}{x}
+ \frac{x}{x^2+1}
- \frac{2}{3(x+1)}\right]
$

What is $\frac{y'}{y}$ ?

 

[MarkFL]

$\displaystyle
\ln y \;=
\;\ln\left[\frac{x(x^2+1)^{\frac{1}{2}}}{(x+1)^{\frac{2}{3}}}\right] \\
=\ln x
+ \frac{1}{2}\ln(x^2+1) - \frac{2}{3}\ln(x+1)$

$\displaystyle
y' =y \left[\frac{1}{x}
+ \frac{x}{x^2+1}
- \frac{2}{3(x+1)}\right]
$

What is $\frac{y'}{y}$ ?

Now that you have solved for $y'$, just substitute for $y$ and you will have the derivative as a function of $x$.

 

[karush]

$\displaystyle \ln y \;=
\;\ln\left[\frac{x(x^2+1)^{\frac{1}{2}}}{(x+1)^{\frac{2}{3}}}\right] \\
=\;\ln x
+ \frac{1}{2}\ln(x^2+1) - \frac{2}{3}\ln(x+1)$

$\displaystyle
y' =\left[\frac{x(x^2+1)^{\frac{1}{2}}}{(x+1)^{\frac{2}{3}}}\right]
\left[\frac{1}{x}
+ \frac{x}{x^2+1}
- \frac{2}{3(x+1)}\right]
$