Question

find dy/dx y = (x^2+1)^10 sec5x​

Answer 1

$y = ( {x}^{2} + 1) ^{10} sec5x \\ \frac{dy}{dx} = ( { {x}^{2} + 1 })^{10} \frac{d}{dx} sec5x + sec5x \frac{d}{dx} ( { {x}^{2} + 1 })^{10} \\ = ( { {x}^{2} + 1})^{10} .sec5xtan5x \frac{d}{dx} 5x + sec5x .\: 10( {x}^{2} + 1) ^{9} \frac{d}{dx} ( {x}^{2} + 1) \\ = 5sec5xtan5x( { {x}^{2} + 1})^{10} + 10( { {x}^{2} + 1 })^{9} 2x.sec5x \\ \frac{dy}{dx} = 5sec5xtan5x( { {x}^{2} + 1})^{10} + 20x( { {x}^{2} + 1})^{9} sec5x$