What is the derivative of g(t)=(9t+5)^2(5t-6)^{-3} ?

The derivative of g(t)=(9t+5)^2(5t-6)^{-3} is (-405t^2-1872t-915)/((5t-6)^4)

What is the first derivative of g(t)=(9t+5)^2(5t-6)^{-3} ?

The first derivative of g(t)=(9t+5)^2(5t-6)^{-3} is (-405t^2-1872t-915)/((5t-6)^4)

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derivative of

g (t) = (9 t + 5)^{2} (5 t - 6)^{- 3}

\frac{- 405 t ^{2} - 1872 t - 915}{( 5 t - 6 ) ^{4}}

Solution steps

\frac{d}{d t} ((9 t + 5)^{2} (5 t - 6)^{- 3})

Apply the Product Rule:

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

f = (9 t + 5)^{2}, g = (5 t - 6)^{- 3}

= \frac{d}{d t} ((9 t + 5)^{2}) (5 t - 6)^{- 3} + \frac{d}{d t} ((5 t - 6)^{- 3}) (9 t + 5)^{2}

\frac{d}{d t} ((9 t + 5)^{2}) = 18 (9 t + 5)

\frac{d}{d t} ((5 t - 6)^{- 3}) = - \frac{15}{( 5 t - 6 ) ^{4}}

= 18 (9 t + 5) (5 t - 6)^{- 3} + (- \frac{15}{( 5 t - 6 ) ^{4}}) (9 t + 5)^{2}

Simplify

18 (9 t + 5) (5 t - 6)^{- 3} + (- \frac{15}{( 5 t - 6 ) ^{4}}) (9 t + 5)^{2} : \frac{- 405 t ^{2} - 1872 t - 915}{( 5 t - 6 ) ^{4}}

= \frac{- 405 t ^{2} - 1872 t - 915}{( 5 t - 6 ) ^{4}}

(3 x + 2) + (3 y - 3) y^{^{'}} = 0

\int y (y + 1) e^{- 2 y} d y

\int e + e^{3 x} d x

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

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

What is the derivative of g(t)=(9t+5)^2(5t-6)^{-3} ?
The derivative of g(t)=(9t+5)^2(5t-6)^{-3} is (-405t^2-1872t-915)/((5t-6)^4)
What is the first derivative of g(t)=(9t+5)^2(5t-6)^{-3} ?
The first derivative of g(t)=(9t+5)^2(5t-6)^{-3} is (-405t^2-1872t-915)/((5t-6)^4)