Advertisements
Advertisements
प्रश्न
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
विकल्प
- \[\frac{1}{3 \ln x}\]
- \[\frac{1}{3 \ln x} - \frac{1}{2 \ln x}\]
(ln x)−1 x (x − 1)
- \[\frac{3 x^2}{\ln x}\]
Advertisements
उत्तर
(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
संबंधित प्रश्न
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
If f(2a − x) = −f(x), prove that
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
