What is the derivative of t/((t-9)^2) ?

The derivative of t/((t-9)^2) is (-9-t)/((t-9)^3)

What is the first derivative of t/((t-9)^2) ?

The first derivative of t/((t-9)^2) is (-9-t)/((t-9)^3)

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derivative of

\frac{t}{( t - 9 ) ^{2}}

\frac{- 9 - t}{( t - 9 ) ^{3}}

Solution steps

\frac{d}{d t} (\frac{t}{( t - 9 ) ^{2}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d t}{d t} ( t - 9 ) ^{2} - \frac{d}{d t} ( ( t - 9 ) ^{2} ) t}{( ( t - 9 ) ^{2} ) ^{2}}

\frac{d t}{d t} = 1

\frac{d}{d t} ((t - 9)^{2}) = 2 (t - 9)

= \frac{1 \cdot ( t - 9 ) ^{2} - 2 ( t - 9 ) t}{( ( t - 9 ) ^{2} ) ^{2}}

Simplify

\frac{1 \cdot ( t - 9 ) ^{2} - 2 ( t - 9 ) t}{( ( t - 9 ) ^{2} ) ^{2}} : \frac{- 9 - t}{( t - 9 ) ^{3}}

= \frac{- 9 - t}{( t - 9 ) ^{3}}

25 y^{^{''}} + 30 y^{^{'}} + 9 y = 0

\frac{\partial}{\partial x} (2 x^{0.3} y^{0.3} + 1)

\int \frac{3 x ^{\frac{8}{3}}}{8} + C d x

\int arccot (5 x) d x

y^{^{'}} + \frac{2}{3} y = 1 - \frac{1}{2} t

What is the derivative of t/((t-9)^2) ?
The derivative of t/((t-9)^2) is (-9-t)/((t-9)^3)
What is the first derivative of t/((t-9)^2) ?
The first derivative of t/((t-9)^2) is (-9-t)/((t-9)^3)