Question

Differentiate the function w.r.t.x
y = (2x +3)(5x^2-7x+1 )​

Answer 1

Given :

y = (2x + 3)( 5x² - 7x + 1)

To find :

Diffrenciation of y with respect to x , i.e., dy/dx

Identity used :

$\sf \: \bullet \: \dfrac{d \: ( x^n)}{dx} \: = \: n \times \bigg( {x}^{(n - 1)} \bigg)$

$\sf \: \bullet \: \dfrac{d \:( a)}{dx} \: = 0 \: \\ \sf \: here \: a \: is \: constant$

Solution :

y = (2x + 3)( 5x² - 7x + 1)

First of all we will find products on RHS

➝ y = (2x)(5x²) - (2x)(7x) + (2x)(1) + (3)(5x²) - (3)(7x) + (3)(1)

➝ y = 10x³ - 14x² + 2x + 15x² - 21x + 3

➝ y = 10x³ + 15x² - 14x² + 2x - 21x + 3

➝ y = 10x³ + x² - 19x + 3

Now we will differentiate y w.r.t. x

$\sf \implies \: \dfrac{dy}{dx} \: = \: \dfrac{d \: ( \: 10 {x}^{3} \: + {x}^{2} - 19x + 3 \: ) }{dx}$

$\sf \implies \: \dfrac{dy}{dx} \: = \: \dfrac{d \: ( \: 10 {x}^{3} \: ) \: }{dx} \: + \: \dfrac{d( {x}^{2}) }{dx} \: - \: \dfrac{d( \: 19x )\: ) }{dx} \: + \: \dfrac{d( \: 3 )\: ) }{dx}$

$\sf \implies \: \dfrac{dy}{dx} \: = \: \bigg(10 \times \dfrac{d \: ( \: {x}^{3} \: ) \: }{dx} \bigg) \: + \: \dfrac{d( {x}^{2}) }{dx} \: - \: \bigg(19 \times \dfrac{d( \: x )\: }{dx} \bigg) \: + \: \dfrac{d( \: 3 )\: }{dx}$

$\sf \implies \: \dfrac{dy}{dx} \: = \: \bigg(10 \times ( \: 3 \times {x}^{(3 - 1)} \: ) \: \bigg) \: + \bigg( 2 \times {x}^{(2 - 1) } \bigg) \: - \: \bigg(19 \times ( \: 1 \times {x}^{(1 - 1)} )\: \bigg) \: + \: 0$

$\sf \implies \: \dfrac{dy}{dx} \: = \: \bigg(10 \times \: 3 \times {x}^{2} \: \: \bigg) \: + \bigg( 2 \times {x}^{1 } \bigg) \: - \: \bigg(19 \times \: 1 \times {x}^{0} \: \bigg) \:$

$\sf \implies \: \dfrac{dy}{dx} \: = \: \bold{30{x}^{2} \: + 2x \: - \: 19 \: }$

________________________________

ANSWER :

Diffrenciation of y w.r.t. x = 30x² + 2x - 19

Answer 2

Step-by-step explanation:

Given :

y = (2x + 3)( 5x² - 7x + 1)

To find :

Diffrenciation of y with respect to x , i.e., dy/dx

Identity used :

\sf \: \bullet \: \dfrac{d \: ( x^n)}{dx} \: = \: n \times \bigg( {x}^{(n - 1)} \bigg)∙

dx

d(x

n

)

=n×(x

(n−1)

)

\begin{gathered}\sf \: \bullet \: \dfrac{d \:( a)}{dx} \: = 0 \: \\ \sf \: here \: a \: is \: constant\end{gathered}

∙

dx

d(a)

=0

hereaisconstant

Solution :

y = (2x + 3)( 5x² - 7x + 1)

First of all we will find products on RHS

➝ y = (2x)(5x²) - (2x)(7x) + (2x)(1) + (3)(5x²) - (3)(7x) + (3)(1)

➝ y = 10x³ - 14x² + 2x + 15x² - 21x + 3

➝ y = 10x³ + 15x² - 14x² + 2x - 21x + 3

➝ y = 10x³ + x² - 19x + 3

Now we will differentiate y w.r.t. x

\sf \implies \: \dfrac{dy}{dx} \: = \: \dfrac{d \: ( \: 10 {x}^{3} \: + {x}^{2} - 19x + 3 \: ) }{dx}⟹

dx

dy

=

dx

d(10x

3

+x

2

−19x+3)

\sf \implies \: \dfrac{dy}{dx} \: = \: \dfrac{d \: ( \: 10 {x}^{3} \: ) \: }{dx} \: + \: \dfrac{d( {x}^{2}) }{dx} \: - \: \dfrac{d( \: 19x )\: ) }{dx} \: + \: \dfrac{d( \: 3 )\: ) }{dx}⟹

dx

dy

=

dx

d(10x

3

)

+

dx

d(x

2

)

−

dx

d(19x))

+

dx

d(3))

\sf \implies \: \dfrac{dy}{dx} \: = \: \bigg(10 \times \dfrac{d \: ( \: {x}^{3} \: ) \: }{dx} \bigg) \: + \: \dfrac{d( {x}^{2}) }{dx} \: - \: \bigg(19 \times \dfrac{d( \: x )\: }{dx} \bigg) \: + \: \dfrac{d( \: 3 )\: }{dx}⟹

dx

dy

=(10×

dx

d(x

3

)

)+

dx

d(x

2

)

−(19×

dx

d(x)

)+

dx

d(3)

\sf \implies \: \dfrac{dy}{dx} \: = \: \bigg(10 \times ( \: 3 \times {x}^{(3 - 1)} \: ) \: \bigg) \: + \bigg( 2 \times {x}^{(2 - 1) } \bigg) \: - \: \bigg(19 \times ( \: 1 \times {x}^{(1 - 1)} )\: \bigg) \: + \: 0⟹

dx

dy

=(10×(3×x

(3−1)

))+(2×x

(2−1)

)−(19×(1×x

(1−1)

))+0

\sf \implies \: \dfrac{dy}{dx} \: = \: \bigg(10 \times \: 3 \times {x}^{2} \: \: \bigg) \: + \bigg( 2 \times {x}^{1 } \bigg) \: - \: \bigg(19 \times \: 1 \times {x}^{0} \: \bigg) \:⟹

dx

dy

=(10×3×x

2

)+(2×x

1

)−(19×1×x

0

)

\sf \implies \: \dfrac{dy}{dx} \: = \: \bold{30{x}^{2} \: + 2x \: - \: 19 \: }⟹

dx

dy

=30x

2

+2x−19

________________________________

ANSWER :

Diffrenciation of y w.r.t. x = 30x² + 2x - 19