Question

Find the derivatives of x^2sec(ax-b)​

Answer 1

Solution:

Given That:

$\rm \longrightarrow y = {x}^{2} \sec(ax - b)$

Taking derivative of both sides wrt x, we get:

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx} \bigg[ {x}^{2} \sec(ax - b) \bigg]$

We know that:

$\bigstar \: \underline{ \boxed{ \rm \dfrac{d}{dx}(fg) =f \dfrac{d}{dx}g + g \dfrac{d}{dx}f }}$

Using this result, we get:

$\rm \longrightarrow \dfrac{dy}{dx} = \sec(ax - b) \cdot \dfrac{d}{dx}{x}^{2} + {x}^{2} \cdot \dfrac{d}{dx} \sec(ax - b)$

$\rm \longrightarrow \dfrac{dy}{dx} = \sec(ax - b) \cdot 2x+ a{x}^{2} \cdot \tan(ax - b) \sec(ax - b)$

$\rm \longrightarrow \dfrac{dy}{dx} = 2x\sec(ax - b) + a{x}^{2} \tan(ax - b) \sec(ax - b)$

$\rm \longrightarrow \dfrac{dy}{dx} = x\sec(ax - b) \big[2 + ax \tan(ax - b) \big]$

★ Which is our required answer.

Learn More:

$\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}$