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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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