Question

differenciate (2x + (1/x))^2

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ \frac{d}{dx} (2x + \frac{1}{x} ) {}^{2} \\ \\ according \: to \: chain \: rule \\ \frac{d}{dx} (u {}^{n} ) = n.u {}^{n - 1} . \frac{d}{dx} (u) \\ \\ = 2.(2x + \frac{1}{x} ) {}^{2 - 1} . \frac{d}{dx} (2x + \frac{1}{x} ) \\ \\we \: know \: that \\ \frac{d}{dx}(x {}^{n} ) = n.x {}^{n - 1} \\ \\ = 2.(2x + \frac{1}{x} ).(2 - \frac{1}{x {}^{2} } )$

Answer 2

Answer:

\begin{gathered}to \: find : \\ \frac{d}{dx} (2x + \frac{1}{x} ) {}^{2} \\ \\ according \: to \: chain \: rule \\ \frac{d}{dx} (u {}^{n} ) = n.u {}^{n - 1} . \frac{d}{dx} (u) \\ \\ = 2.(2x + \frac{1}{x} ) {}^{2 - 1} . \frac{d}{dx} (2x + \frac{1}{x} ) \\ \\we \: know \: that \\ \frac{d}{dx}(x {}^{n} ) = n.x {}^{n - 1} \\ \\ = 2.(2x + \frac{1}{x} ).(2 - \frac{1}{x {}^{2} } )\end{gathered}

tofind:

dx

d

(2x+

x

1

)

2

accordingtochainrule

dx

d

(u

n

)=n.u

n−1

.

dx

d

(u)

=2.(2x+

x

1

)

2−1

.

dx

d

(2x+

x

1

)

weknowthat

dx

d

(x

n

)=n.x

n−1

=2.(2x+

x

1

).(2−

x

2

1

)

Hope it helps u mark me brainlest