What is the derivative of y=(88t)/(99t-88) ?

The derivative of y=(88t)/(99t-88) is -(64)/((9t-8)^2)

What is the first derivative of y=(88t)/(99t-88) ?

The first derivative of y=(88t)/(99t-88) is -(64)/((9t-8)^2)

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derivative of

y = \frac{88 t}{99 t - 88}

- \frac{64}{( 9 t - 8 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{88 t}{99 t - 88})

Take the constant out:

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

= 88 \frac{d}{d t} (\frac{t}{99 t - 88})

Apply the Quotient Rule:

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

= 88 \cdot \frac{\frac{d t}{d t} ( 99 t - 88 ) - \frac{d}{d t} ( 99 t - 88 ) t}{( 99 t - 88 ) ^{2}}

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

\frac{d}{d t} (99 t - 88) = 99

= 88 \cdot \frac{1 \cdot ( 99 t - 88 ) - 99 t}{( 99 t - 88 ) ^{2}}

Simplify

88 \cdot \frac{1 \cdot ( 99 t - 88 ) - 99 t}{( 99 t - 88 ) ^{2}} : - \frac{64}{( 9 t - 8 ) ^{2}}

= - \frac{64}{( 9 t - 8 ) ^{2}}

\frac{d}{d x} (x - y^{2})

\frac{d}{d x} (x^{\frac{2}{3}} (x - 2))

y^{^{'}} = \frac{3 x ^{2} + y ^{2}}{x y}

ln (sec (2 x) + tan (2 x))

(- 3, - 9), (8, 3)

What is the derivative of y=(88t)/(99t-88) ?
The derivative of y=(88t)/(99t-88) is -(64)/((9t-8)^2)
What is the first derivative of y=(88t)/(99t-88) ?
The first derivative of y=(88t)/(99t-88) is -(64)/((9t-8)^2)