Question

d/dx[x^2 secx]=
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Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \dfrac{d}{dx} ({x}^{2} secx)$

We know,

$\boxed{\rm{  \:\dfrac{d}{dx}u.v \: = \: u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \: \: }}$

So, using this result, we get

$\rm \: =  \: {x}^{2}\dfrac{d}{dx}secx + secx\dfrac{d}{dx} {x}^{2}$

We know,

$\boxed{\rm{  \:\dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: \: }}$

and

$\boxed{\rm{  \:\dfrac{d}{dx}secx \: = \: secx \: tanx \: \: }}$

So, using these results, we get

$\rm \: =  \: {x}^{2}(secx \: tanx) + secx \: (2x)$

$\rm \: =  \:x \: secx \: (x \: tanx \: + \: 2)$

Hence,

$\rm\implies \:\boxed{\rm{  \:\dfrac{d}{dx} ({x}^{2}secx) =  \:x \: secx \: (x \: tanx \: + \: 2) \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

• Possible derivation:

$\sf \[ \dfrac{d}{d x}\left(x^{2} \sec (x)\right) \]$

• Use the product rule,

$\tt \dfrac{d}{d x}(u v)=v \dfrac{d u}{d x}+u \dfrac{d v}{d x}$, where $\tt u=x^{2}$ and $\tt v=\sec (x):$

$\sf =x^{2}\left(\dfrac{d}{d x}\sec (x)\right)+\left(\dfrac{d}{d x}\left(x^{2}\right)\right) \sec (x)$

• Using the chain rule,

$\tt \dfrac{d}{d x}(\sec (x))=\dfrac{d \sec (u)}{d u} \dfrac{d u}{d x}$,where u=x and

$\tt \dfrac{d}{d u}(\sec (u))=\sec (u)\tan (u):$

$\sf=\left(\dfrac{d}{d x}\left(x^{2}\right)\right) \sec (x)+ \boxed{ \sf\left(\dfrac{d}{d x}(x)\right) \sec (x) \tan (x)} x^{2}$

• The derivative of x is 1 :

$\sf\[ =\left(\dfrac{d}{d x}\left(x^{2}\right)\right) \sec (x)+ \boxed{1} \: \: x^{2} \sec (x) \tan (x) \]$

• Use the power rule, $\tt \dfrac{d}{d x}\left(x^{n}\right)=n x^{n-1}$,
• where n=2 .

$\[ \begin{array}{l} \sf \dfrac{d}{d x}\left(x^{2}\right)= 2 x: \\ \\ \sf = \boxed{ \sf2 x } \: \sec (x)+x^{2} \sec (x) \tan (x) \end{array} \]$

• Simplify the expression:

$\colorbox{powderblue}{ \boxed{ \begin{array}{l} \underline{\underline{ \color{orange} \text{Answer: }}} \\ \\ & \color{red} \[ \sf =x \sec (x)(2+x \tan (x)) \] \end{array}}}$