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Answers

Answered by yj77443
1

Explanation:

Given :

y = 5(x² + 3x + 2)³

To Find :

\sf{\dfrac{dy}{dx}}

dx

dy

Solution :

Let x² + 3x + 2 = u. Then y = 5u³

• Differentiating y with respect to u ;

\begin{gathered} \\ : \implies \bf \: \frac{dy}{du} = \frac{d}{du} (5u {}^{3} ) \\ \end{gathered}

:⟹

du

dy

=

du

d

(5u

3

)

• By applying , \bf{\dfrac{d}{dx}(x^n)=\dfrac{d}{dx}(nx^{n-1})}

dx

d

(x

n

)=

dx

d

(nx

n−1

) . We get ;

\begin{gathered} \\ : \implies \bf \: \frac{dy}{du} = (15u {}^{2} ) \sf \: ..........(1)\end{gathered}

:⟹

du

dy

=(15u

2

)..........(1)

Now , differentiating u with respect to x ;

\begin{gathered} \\ : \implies \bf \: \frac{du}{dx} = \frac{d}{dx} ( {x}^{2} + 3x + 2) \\ \end{gathered}

:⟹

dx

du

=

dx

d

(x

2

+3x+2)

\begin{gathered} \\ : \implies \bf \: \frac{du}{dx} = \frac{d}{dx} ( {x}^{2} ) + \frac{d}{dx} (3x) + \frac{d}{dx} (3) \\ \end{gathered}

:⟹

dx

du

=

dx

d

(x

2

)+

dx

d

(3x)+

dx

d

(3)

• By applying , \bf{\dfrac{d}{dx}(x^n)=\dfrac{d}{dx}(nx^{n-1})}

dx

d

(x

n

)=

dx

d

(nx

n−1

) . We get ;

\begin{gathered} \\ : \implies \bf \: \frac{du}{dx} = \frac{d}{dx} (2 {x}^{2 - 1} ) + \frac{d}{dx} (3 \times 1 {x}^{1 - 1} ) + \frac{d}{dx} (2) \\ \end{gathered}

:⟹

dx

du

=

dx

d

(2x

2−1

)+

dx

d

(3×1x

1−1

)+

dx

d

(2)

\begin{gathered} \\ : \implies \bf \frac{du}{dx} = 2x + 3 \times 1 {x}^{0} + \frac{d}{dx} (2) \\ \end{gathered}

:⟹

dx

du

=2x+3×1x

0

+

dx

d

(2)

• We know that a⁰ = 1. ;

\begin{gathered} \\ : \implies \bf \: \frac{du}{dx} = 2x + 3(1) + \frac{d}{dx}(2) \\ \end{gathered}

:⟹

dx

du

=2x+3(1)+

dx

d

(2)

\begin{gathered} \\ : \implies \bf \: \frac{du}{dx} = 2x + 3 + \frac{d}{dx} (2) \\ \end{gathered}

:⟹

dx

du

=2x+3+

dx

d

(2)

• Differentitation of a Constant is "zero".

\begin{gathered} \\ : \implies \bf \: \frac{du}{dx} = 2x + 3 + 0 \\ \end{gathered}

:⟹

dx

du

=2x+3+0

\begin{gathered} \\ : \implies \bf \frac{du}{dx} = 2x + 3 \sf \: ...........(2)\\ \end{gathered}

:⟹

dx

du

=2x+3...........(2)

• Now by applying chain rule i.e ,

\begin{gathered} \\ : \implies \bf \: \frac{dy}{du}.\frac{du}{dx} = \frac{dy}{dx} \end{gathered}

:⟹

du

dy

.

dx

du

=

dx

dy

• From eq(1) & eq(2) . We get ;

\begin{gathered} \\ : \implies \bf \: \frac{dy}{dx} = 15 {u}^{2} \times (2x + 3) \\ \end{gathered}

:⟹

dx

dy

=15u

2

×(2x+3)

• Since u = x² + 3x + 2 ;

\begin{gathered} \\ : \implies {\underline{\boxed{\bf \: \frac{dy}{dx} = 15( {x}^{2} + 3x + 2) {}^{2} \times 2x + 3}}}\end{gathered}

:⟹

dx

dy

=15(x

2

+3x+2)

2

×2x+3

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