What is the derivative of y= t/((t-4)^2) ?

The derivative of y= t/((t-4)^2) is (-4-t)/((t-4)^3)

What is the first derivative of y= t/((t-4)^2) ?

The first derivative of y= t/((t-4)^2) is (-4-t)/((t-4)^3)

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derivative of

y = \frac{t}{( t - 4 ) ^{2}}

\frac{- 4 - t}{( t - 4 ) ^{3}}

Solution steps

\frac{d}{d t} (\frac{t}{( t - 4 ) ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d t}{d t} ( t - 4 ) ^{2} - \frac{d}{d t} ( ( t - 4 ) ^{2} ) t}{( ( t - 4 ) ^{2} ) ^{2}}

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

\frac{d}{d t} ((t - 4)^{2}) = 2 (t - 4)

= \frac{1 \cdot ( t - 4 ) ^{2} - 2 ( t - 4 ) t}{( ( t - 4 ) ^{2} ) ^{2}}

Simplify

\frac{1 \cdot ( t - 4 ) ^{2} - 2 ( t - 4 ) t}{( ( t - 4 ) ^{2} ) ^{2}} : \frac{- 4 - t}{( t - 4 ) ^{3}}

= \frac{- 4 - t}{( t - 4 ) ^{3}}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = sin (x)

n = 0 \sum \infty \frac{1}{( 4 n + 3 ) ^{2}}

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

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

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

What is the derivative of y= t/((t-4)^2) ?
The derivative of y= t/((t-4)^2) is (-4-t)/((t-4)^3)
What is the first derivative of y= t/((t-4)^2) ?
The first derivative of y= t/((t-4)^2) is (-4-t)/((t-4)^3)