What is the derivative of regardsp I/(2p) ?

The derivative of regardsp I/(2p) is ergaIds

What is the first derivative of regardsp I/(2p) ?

The first derivative of regardsp I/(2p) is ergaIds

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derivative of

re g a r d s p \frac{I}{2 p}

er g a I d s

Solution steps

\frac{d}{d r} (re g a r d s p \frac{I}{2 p})

Treat

g, a, s, p, I

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= e g a d s p \frac{I}{2 p} \frac{d}{d r} (rr)

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = r, g = r

= e g a d s p \frac{I}{2 p} (\frac{d r}{d r} r + \frac{d r}{d r} r)

\frac{d r}{d r} = 1

\frac{d r}{d r} = 1

= e g a d s p \frac{I}{2 p} (1 \cdot r + 1 \cdot r)

Simplify

e g a d s p \frac{I}{2 p} (1 \cdot r + 1 \cdot r) : er g a I d s

= er g a I d s

\int \frac{1}{4 ( 1 - x ) ^{2}} d x

\frac{d ^{2} y}{d x ^{2}} = 4 x

\int \frac{6 x ^{2} + x + 6}{( x ^{2} + 1 ) ^{2}} d x

ln (x - 4)

\frac{d}{d x} (ln (sin (4 x^{3} + 6 x^{2} + 8)))

What is the derivative of regardsp I/(2p) ?
The derivative of regardsp I/(2p) is ergaIds
What is the first derivative of regardsp I/(2p) ?
The first derivative of regardsp I/(2p) is ergaIds