Question

differentiate w.r.to x : x(x-3) (x^2+x)​

Answer 1

Solution:

Given That:

$\rm \longrightarrow y =x(x - 3)( {x}^{2} + x)$

$\rm \longrightarrow y =( {x}^{2} - 3x)( {x}^{2} + x)$

$\rm \longrightarrow y = {x}^{4} + {x}^{3} - 3 {x}^{3} - 3x$

$\rm \longrightarrow y = {x}^{4} - 2 {x}^{3} -3 {x}^{2}$

Differentiating both sides wrt x, we get:

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx}( {x}^{4} - 2{x}^{3} - 3{x}^{2})$

We know that:

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

Using this result, we get:

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx}( {x}^{4}) - \dfrac{d}{dx}( 2 {x}^{3}) - \dfrac{d}{dx} ( 3{x}^{2})$

We know that:

$\bigstar \: \underline{ \boxed{ \rm \dfrac{d}{dx}( {x}^{n} ) =n {x}^{n- 1} }}$

$\bigstar \: \underline{ \boxed{ \rm \dfrac{d}{dx}(C \cdot f) =C \dfrac{d}{dx}f }}$

Using this result, we get:

$\rm \longrightarrow \dfrac{dy}{dx} =4 {x}^{3} - 2\times 3 \times {x}^{2} - 3 \times 2 \times x$

$\rm \longrightarrow \dfrac{dy}{dx} =4 {x}^{3} -6{x}^{2} - 6x$

★ Which is our required answer.

Learn More:

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