Question

(x-5)(x-3) differentiate with respect to x​

Answer 1

Given

( x - 5 ) ( x - 3 )

To Find

To differentiate the given equation with respect to x

Formula Applied

$\bullet\ \; \rm \dfrac{d}{dx}(x^n)=nx^{n-1}\\\\\bullet\ \; \rm \dfrac{d}{dx}(kx^n)=k\dfrac{d}{dx}(x^n)$

where , k is constant

$\bullet\ \; \rm \dfrac{d}{dx}(k)=0$

Proof :

$\rm \dfrac{d}{dx}(k)=0\\\\\to \rm k\dfrac{d}{dx}(x^0)=0\\\\\to \rm k[0(x^{0-1})]=0\\\\\to \rm k(0)=0\\\\\to \bf 0=0$

Solution

Let ,

y = ( x - 5 ) ( x - 3 )

⇒ y = x² - 3x - 5x + 15

⇒ y = x² - 8x + 15

Differentiate with respect to x on both sides ,

$\implies \rm \dfrac{dy}{dx}=\dfrac{d}{dx}(x^2-8x+15)\\\\\to \rm \dfrac{dy}{dx}=\dfrac{d}{dx}(x^2)-\dfrac{d}{dx}(8x)+\dfrac{d}{dx}(15)\\\\\to \rm \dfrac{dy}{dx}=2x-8+0\\\\\bf \to \pink{\dfrac{dy}{dx}=2x-8\ \; \bigstar}$

Answer 2

Answer:

Given

( x - 5 ) ( x - 3 )

To Find

To differentiate the given equation with respect to x

Formula Applied

$\begin{gathered}\bullet\ \; \rm \dfrac{d}{dx}(x^n)=nx^{n-1}\\\\\bullet\ \; \rm \dfrac{d}{dx}(kx^n)=k\dfrac{d}{dx}(x^n)\end{gathered} </p><p>∙$

where , k is constant

$\bullet\ \; \rm \dfrac{d}{dx}(k)=0∙ </p><p>$

Proof :

$\begin{gathered}\rm \dfrac{d}{dx}(k)=0\\\\\to \rm k\dfrac \pink{d} \pink{dx}(x^0)=0\\\\\to \rm k[0(x^{0-1})]=0\\\\\to \rm k(0)=0\\\\\to \bf 0=0\end{gathered} </p><p></p><p>$

Solution

Let ,

y = ( x - 5 ) ( x - 3 )

⇒ y = x² - 3x - 5x + 15

⇒ y = x² - 8x + 15

Differentiate with respect to x on both sides ,

$\begin{gathered}\implies \rm \dfrac{dy}{dx}=\dfrac{d}{dx}(x^2-8x+15)\\\\\to \rm \dfrac{dy}{dx}=\dfrac{d}{dx}(x^2)-\dfrac{d}{dx}(8x)+\dfrac{d}{dx}(15)\\\\\to \rm \dfrac{dy}{dx}=2x-8+0\\\\\bf \to \red{\dfrac{dy}{dx}=2x-8\ \; \bigstar}\end{gathered}$