differentiate the function (x-1/x)²
Answers
Step-by-step explanation:
We have to differentiate the given function :-
\rm y = {\bigg (x - \dfrac{1}{x} \bigg)}^{2}y=(x−
x
1
)
2
\begin{gathered}\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d{\bigg (x - \dfrac{1}{x} \bigg)}^{2} }{dx} \\ \\ \\ \rm \longrightarrow \dfrac{dy}{dx} = 2 \times \bigg(x - \dfrac{1}{x} \bigg) \times \dfrac{d{\bigg (x - \dfrac{1}{x} \bigg)} }{dx}\\ \\ \\ \rm \longrightarrow \dfrac{dy}{dx} = 2 \times \bigg(x - \dfrac{1}{x} \bigg) \times \bigg ( \dfrac{d{ (x )} }{dx} - \dfrac{d{ ( \frac{1}{x} )} }{dx}\bigg)\\ \\ \\ \rm \longrightarrow \dfrac{dy}{dx} = 2 \times \bigg(x - \dfrac{1}{x} \bigg) \times \bigg (1- \dfrac{d{ ( {x}^{{ - 1}} )} }{dx}\bigg)\\ \\ \\ \rm \longrightarrow \dfrac{dy}{dx} = 2 \times \bigg(x - \dfrac{1}{x} \bigg) \times \bigg (1- {( - {x}^{ - 2} ) }\bigg)\\ \\ \\ \rm \longrightarrow \dfrac{dy}{dx} = 2 \times \bigg(x - \dfrac{1}{x} \bigg) \times \bigg (1 + {x}^{ - 2}\bigg)\\ \\ \\ \rm \longrightarrow \dfrac{dy}{dx} = 2 \times \bigg(x - \dfrac{1}{x} \bigg) \times \bigg (1 + \dfrac{1}{ {x}^{2} }\bigg)\\ \\ \\ \rm \longrightarrow \dfrac{dy}{dx} = 2 \times \bigg(x + \dfrac{1}{x} - \dfrac{1}{x} -\dfrac{1}{ {x}^{3} } \bigg)\\ \\ \\ \rm \longrightarrow \dfrac{dy}{dx} = 2 \times \bigg(x -\dfrac{1}{ {x}^{3} } \bigg)\\ \\ \\ \rm \longrightarrow \dfrac{dy}{dx} = \dfrac{ 2{x}^{4} - 2}{ {x}^{3} } \end{gathered}
⟶
dx
dy
=
dx
d(x−
x
1
)
2
⟶
dx
dy
=2×(x−
x
1
)×
dx
d(x−
x
1
)
⟶
dx
dy
=2×(x−
x
1
)×(
dx
d(x)
−
dx
d(
x
1
)
)
⟶
dx
dy
=2×(x−
x
1
)×(1−
dx
d(x
−1
)
)
⟶
dx
dy
=2×(x−
x
1
)×(1−(−x
−2
))
⟶
dx
dy
=2×(x−
x
1
)×(1+x
−2
)
⟶
dx
dy
=2×(x−
x
1
)×(1+
x
2
1
)
⟶
dx
dy
=2×(x+
x
1
−
x
1
−
x
3
1
)
⟶
dx
dy
=2×(x−
x
3
1
)
⟶
dx
dy
=
x
3
2x
4
−2
f(x) = (x - 1/x)²
f' = 2(x - 1/x) × d/dx (x - 1/x) [ Chain rule ]
=> f' = 2(x - 1/x) × [1 - (- 1/x²) ]
=> f' = 2(x - 1/x)( 1 + 1/x²)