# Thread: need help for differentiation problem

1. ## need help for differentiation problem

differentiate: y = x^ ½ (1 – 2x)^ 2/3 ÷ (2 – 3x)^ 3/4 (3 – 4x)^ 4/5 x^1/2 means x power 1/2 this is how i have started, but im stuck.. is this the right way of doing it?plz help me ln y = (½ ln x)(2/3 ln (1 – 2x)) – ¾ ln (2 -3x) 4/5 ln (3 -4x)

2. Is this the problem:

$\displaystyle \frac{x^{\frac{1}{2}}(1-2x)^{\frac{2}{3}}}{(2-3x)^{\frac{3}{4}}(3-4x)^{\frac{4}{5}}x^{\frac{1}{2}}}$

If so, first notice the $\displaystyle x^{1/2}$ cancels.

You could use the product and quotient rules.

3. ## Re:differentiation problem

no. the problem is
[IMG]file:///C:/DOCUME%7E1/yajna10/LOCALS%7E1/Temp/moz-screenshot-2.jpg[/IMG] y = x^1/2(1-2x)^2/3 ÷ (2-3x)^3/4(3-4x)^4/5

can i use ln first then differentiate..
can you show me how to do it plz??

4. Hello, slash!

Differentiate: .$\displaystyle y \;= \;\frac{x^{\frac{1}{2}}(1 - 2x)^{\frac{2}{3}}}{(2 - 3x)^{\frac{3}{4}}(3 - 4x)^{\frac{4}{5}}}$
With such an ugly function, log differentiation is probably expected.

Take logs: .$\displaystyle \ln(y) \;=\;\ln\left[\frac{x^{\frac{1}{2}}(1-2x)^{\frac{2}{3}}}{(2-3x)^{\frac{3}{4}}(3-4x)^{\frac{4}{5}}}\right]$

. . $\displaystyle \ln(y) \;=\;\ln\left(x^{\frac{1}{2}}\right) - \ln(1-2x)^{\frac{2}{3}} - \ln(2-3x)^{\frac{3}{4}} - \ln(3-4x)^{\frac{4}{5}}$

. . $\displaystyle \ln(y) \;=\;\frac{1}{2}\ln(x) + \frac{2}{3}\ln(1-2x) - \frac{3}{4}\ln(2-3x) - \frac{4}{5}\ln(3-4x)$

Differentiate implicitly: .$\displaystyle \frac{1}{y}\!\cdot\!\frac{dy}{dx}\;=\;\frac{1}{2}\ !\cdot\!\frac{1}{x} + \frac{2}{3}\!\cdot\!\frac{\text{-}2}{1-2x} - \frac{3}{4}\!\cdot\!\frac{\text{-}3}{2-3x} - \frac{4}{5}\!\cdot\!\frac{\text{-}4}{3-4x}$

. . $\displaystyle \frac{1}{y}\cdot\frac{dy}{dx}\;=\;\frac{1}{2x} - \frac{4}{3(1-2x)} + \frac{9}{4(2-3x)} + \frac{16}{5(3-4x)}$

Then: .$\displaystyle \frac{dy}{dx} \;=\;y\left[\frac{1}{2x} - \frac{4}{3(1-2x)} + \frac{9}{4(2-3x)} + \frac{16}{5(3-4x)}\right]$

Therefore: .$\displaystyle \frac{dy}{dx} \;\;=\;\;\frac{x^{\frac{1}{2}}(1-2x)^{\frac{2}{3}}}{(2-3x)^{\frac{3}{4}}(3-4x)^{\frac{4}{5}}} \cdot\left[\frac{1}{2x} - \frac{4}{3(1-2x)} + \frac{9}{4(2-3x)} + \frac{16}{5(3-4x)}\right]$

5. Originally Posted by Soroban
Hello, slash!

With such an ugly function, log differentiation is probably expected.

Take logs: .$\displaystyle \ln(y) \;=\;\ln\left[\frac{x^{\frac{1}{2}}(1-2x)^{\frac{2}{3}}}{(2-3x)^{\frac{3}{4}}(3-4x)^{\frac{4}{5}}}\right]$

. . $\displaystyle \ln(y) \;=\;\ln\left(x^{\frac{1}{2}}\right) - \ln(1-2x)^{\frac{2}{3}} - \ln(2-3x)^{\frac{3}{4}} - \ln(3-4x)^{\frac{4}{5}}$

. . $\displaystyle \ln(y) \;=\;\frac{1}{2}\ln(x) + \frac{2}{3}\ln(1-2x) - \frac{3}{4}\ln(2-3x) - \frac{4}{5}\ln(3-4x)$

Differentiate implicitly: .$\displaystyle \frac{1}{y}\!\cdot\!\frac{dy}{dx}\;=\;\frac{1}{2}\ !\cdot\!\frac{1}{x} + \frac{2}{3}\!\cdot\!\frac{\text{-}2}{1-2x} - \frac{3}{4}\!\cdot\!\frac{\text{-}3}{2-3x} - \frac{4}{5}\!\cdot\!\frac{\text{-}4}{3-4x}$

. . $\displaystyle \frac{1}{y}\cdot\frac{dy}{dx}\;=\;\frac{1}{2x} - \frac{4}{3(1-2x)} + \frac{9}{4(2-3x)} + \frac{16}{5(3-4x)}$

Then: .$\displaystyle \frac{dy}{dx} \;=\;y\left[\frac{1}{2x} - \frac{4}{3(1-2x)} + \frac{9}{4(2-3x)} + \frac{16}{5(3-4x)}\right]$

Therefore: .$\displaystyle \frac{dy}{dx} \;\;=\;\;\frac{x^{\frac{1}{2}}(1-2x)^{\frac{2}{3}}}{(2-3x)^{\frac{3}{4}}(3-4x)^{\frac{4}{5}}} \cdot\left[\frac{1}{2x} - \frac{4}{3(1-2x)} + \frac{9}{4(2-3x)} + \frac{16}{5(3-4x)}\right]$

can you explain me the first line of your answer plz..why it is ln x^1/2 - ln (1-2x)^2/3 rather than (ln x^1/2)X(ln (1-2x)^2/3)

6. Log. of a product

$\displaystyle \ln (a\cdot b)=\ln a+\ln b,\,\forall a,b\in\mathbb R^+$