Question

differentiate
$({x}^{3} + 1)(3 {x}^{2} + 2x - 7)$
by differentiating with respect to x​

Answer 1

Solution:

Given That:

$\rm \longrightarrow y =( {x}^{3} + 1)(3 {x}^{2} + 2x - 7)$

$\rm \longrightarrow y =3 {x}^{5} + 2 {x}^{4} - 7 {x}^{3} +3 {x}^{2} + 2x - 7$

Differentiating both sides wrt x, we get:

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx}(3 {x}^{5} + 2 {x}^{4} - 7 {x}^{3} +3 {x}^{2} + 2x - 7)$

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:

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx}(3 {x}^{5}) + \dfrac{d}{dx} ( 2 {x}^{4}) - \dfrac{d}{dx}(7 {x}^{3})+ \dfrac{d}{dx}(3 {x}^{2}) + \dfrac{d}{dx}( 2x )- \dfrac{d}{dx} (7)$

We know that:

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

Using this result, we get:

$\rm \longrightarrow \dfrac{dy}{dx} = 3\dfrac{d}{dx}({x}^{5}) + 2\dfrac{d}{dx} ( {x}^{4}) - 7\dfrac{d}{dx}({x}^{3})+ 3\dfrac{d}{dx}({x}^{2}) + 2\dfrac{d}{dx}(x)- \dfrac{d}{dx} (7)$

We know that:

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

$\bigstar \: \underline{ \boxed{ \rm \dfrac{d}{dx}( C ) =0}}$

Using this result, we get:

$\rm \longrightarrow \dfrac{dy}{dx} = (3 \cdot 5 \cdot{x}^{4})+ (2 \cdot4 \cdot {x}^{3} ) - (7 \cdot3 \cdot {x}^{2})+ (3 \cdot2 \cdot x )+ (2 \times 1)-0$

$\rm \longrightarrow \dfrac{dy}{dx} =15{x}^{4}+8{x}^{3} -21{x}^{2} + 6x+2$

★ Which is our required answer.

Learn More:

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