What is derivative of g(y)=-y/((y^2+1)^{3/2)} ?

The answer to derivative of g(y)=-y/((y^2+1)^{3/2)} is (dy)/(dx)=0

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derivative of g(y)=-y/((y^2+1)^{3/2)}

derivative of

g (y) = - \frac{y}{( y ^{2} + 1 ) ^{\frac{3}{2}}}

\frac{d y}{d x} = 0

Solution steps

g (y) = - \frac{y}{( y ^{2} + 1 ) ^{\frac{3}{2}}}

Treat

y

y (x)

Differentiate both sides:

0 = - \frac{\frac{d y}{d x} ( - 2 y ^{2} + 1 )}{( y ^{2} + 1 ) ^{\frac{5}{2}}}

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = 0

\frac{d y}{d x} = 0

\int ln (x^{2} + 15 x + 44) d x

\int \frac{1}{( 484 + x ^{2} ) ^{\frac{3}{2}}} d x

\frac{3 ^{2}}{( s + 3 ) ^{2}} \cdot \frac{1}{s}

\int \frac{1 - w}{w} d w

f (x) = sin^{2} (4 x)

What is derivative of g(y)=-y/((y^2+1)^{3/2)} ?
The answer to derivative of g(y)=-y/((y^2+1)^{3/2)} is (dy)/(dx)=0