Advertisements
Advertisements
Question
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
Options
- \[\frac{1}{3 \ln x}\]
- \[\frac{1}{3 \ln x} - \frac{1}{2 \ln x}\]
(ln x)−1 x (x − 1)
- \[\frac{3 x^2}{\ln x}\]
Advertisements
Solution
(ln x)−1 x (x − 1)
Using Newton Leibnitz formula
\[f' (x) = \frac{1}{\log_e x^3}(3 x^2 ) - \frac{1}{\log_e x^2}(2x) \]
\[= \frac{3 x^2}{3\ln x}- \frac{2x}{2\ln x} \]
\[= \frac{x^2}{\ln x} - \frac{x}{\ln x} \]
\[= \frac{1}{\ln x}x(x - 1) \]
\[= {(\ln x)}^{- 1} x(x - 1)\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_2^3 e^{- x} dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
