For Chinu - (2)

Calculus Level 2

$\Large \dfrac{d}{dx} \int_{1}^{x^4} \sec t \ dt = \ ?$

5x^4\tan x^5

4x^3\tan x^4

4x^3\sec x^4

5x^4\sec x^5

1 solution

Aditya Kumar
Feb 5, 2016

By Leibniz rule we have: $\frac { d }{ dx } \int _{ g\left( x \right) }^{ f\left( x \right) }{ h\left( t \right) dt } =f'\left( x \right) h\left( f\left( x \right) \right) -g'\left( x \right) h\left( g\left( x \right) \right)$

Here $f\left( x \right) ={ x }^{ 4 },\quad g\left( x \right) =1,\quad h\left( t \right) =\sec \left( t \right)$

Therefore, $\displaystyle \frac { d }{ dx } \int _{ 1 }^{ { x }^{ 4 } }{ \sec \left( t \right) dt } =4{ x }^{ 3 }\sec \left( { x }^{ 4 } \right)$

For Chinu - (2)

1 solution

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