Question

Using logarithmic differentiation to find the derivative of f(x)=[(e^3x√2x-5)/(6-5x)^4]


(2) Find the integral of[ (x^3+1)^1/3x
^5]dx

Answer 1

Given :

$\frac{(e^{3x})(\sqrt{2x-5)}}{(6-5x)^{4}}=y$

Rewrite as ${(lne^{3x})+ln\sqrt{(2x-5)}}-{ln(6-5x)^{4}}=lny$

${(3x})+{\frac{1}{2}*{ln(2x-5)}}-{4ln (6-5x)}=ln y$

Differentiate

3+$\frac{1}{2x-5}-\frac{4(-5)}{6-5x} = \frac{1 y'}{y}$

$y'= y(3+\frac{1}{2x-5}+\frac{20}{6-5x})$

Solution : y' = e^3x$√2x-5)/(6-5x)^4](3+\frac{1}{2x-5}+\frac{20}{6-5x})$

Part 2:

To find : $\int\frac{\left(1+x^{\left(3\right)^{ }}\right)^{\frac{1}{3}}}{x^{5}} dx$

solution : Take u= $\frac{\left(1+x^{\left(3\right)^{ }}\right)^{\frac{1}{3}}}{x^{ }}$

du= $\frac{x}{\left(1+x^{\left(3\right)}\right)^{\frac{2}{3}}}-\frac{\left(1+x^{\left(3\right)^{ }}\right)^{\frac{1}{3}}}{x^{2}}$ dx

Hence $\int\frac{\left(1+x^{\left(3\right)^{ }}\right)^{\frac{1}{3}}}{x^{5}} dx$ = $\int$ -u³ du

= -$\frac{u^{4}}{4} +c$

substitute for u , and the solution is =- $\frac{(x^{3}+1)^{\frac{4}{3}}}{4x^{4}} +c$