What is the d/(dy)(2kqy(y^2+d^2)^{-3/2}) ?

The d/(dy)(2kqy(y^2+d^2)^{-3/2}) is (2kq(-2y^2+d^2))/((y^2+d^2)^{5/2)}

What is the first d/(dy)(2kqy(y^2+d^2)^{-3/2}) ?

The first d/(dy)(2kqy(y^2+d^2)^{-3/2}) is (2kq(-2y^2+d^2))/((y^2+d^2)^{5/2)}

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\frac{d}{d y} (2 k q y (y^{2} + d^{2})^{- \frac{3}{2}})

\frac{2 k q ( - 2 y ^{2} + d ^{2} )}{( y ^{2} + d ^{2} ) ^{\frac{5}{2}}}

Solution steps

\frac{d}{d y} (2 k q y (y^{2} + d^{2})^{- \frac{3}{2}})

Treat

d, k, q

as a constant

Take the constant out:

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

= 2 k q \frac{d}{d y} (y (y^{2} + d^{2})^{- \frac{3}{2}})

Apply the Product Rule:

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

f = y, g = (y^{2} + d^{2})^{- \frac{3}{2}}

= 2 k q (\frac{d y}{d y} (y^{2} + d^{2})^{- \frac{3}{2}} + \frac{d}{d y} ((y^{2} + d^{2})^{- \frac{3}{2}}) y)

\frac{d y}{d y} = 1

\frac{d}{d y} ((y^{2} + d^{2})^{- \frac{3}{2}}) = - \frac{3 y}{( y ^{2} + d ^{2} ) ^{\frac{5}{2}}}

= 2 k q (1 \cdot (y^{2} + d^{2})^{- \frac{3}{2}} + (- \frac{3 y}{( y ^{2} + d ^{2} ) ^{\frac{5}{2}}}) y)

Simplify

2 k q (1 \cdot (y^{2} + d^{2})^{- \frac{3}{2}} + (- \frac{3 y}{( y ^{2} + d ^{2} ) ^{\frac{5}{2}}}) y) : \frac{2 k q ( - 2 y ^{2} + d ^{2} )}{( y ^{2} + d ^{2} ) ^{\frac{5}{2}}}

= \frac{2 k q ( - 2 y ^{2} + d ^{2} )}{( y ^{2} + d ^{2} ) ^{\frac{5}{2}}}

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What is the d/(dy)(2kqy(y^2+d^2)^{-3/2}) ?
The d/(dy)(2kqy(y^2+d^2)^{-3/2}) is (2kq(-2y^2+d^2))/((y^2+d^2)^{5/2)}
What is the first d/(dy)(2kqy(y^2+d^2)^{-3/2}) ?
The first d/(dy)(2kqy(y^2+d^2)^{-3/2}) is (2kq(-2y^2+d^2))/((y^2+d^2)^{5/2)}