What is the derivative of f(t)=9tsin(pit) ?

The derivative of f(t)=9tsin(pit) is 9(sin(pit)+pitcos(pit))

What is the first derivative of f(t)=9tsin(pit) ?

The first derivative of f(t)=9tsin(pit) is 9(sin(pit)+pitcos(pit))

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derivative of

f (t) = 9 t sin (π t)

9 (sin (π t) + π t cos (π t))

Solution steps

\frac{d}{d t} (9 t sin (π t))

Take the constant out:

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

= 9 \frac{d}{d t} (t sin (π t))

Apply the Product Rule:

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

f = t, g = sin (π t)

= 9 (\frac{d t}{d t} sin (π t) + \frac{d}{d t} (sin (π t)) t)

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

\frac{d}{d t} (sin (π t)) = cos (π t) π

= 9 (1 \cdot sin (π t) + cos (π t) π t)

Simplify

= 9 (sin (π t) + π t cos (π t))

\frac{\partial}{\partial x} (1000 e^{- x^{2} - 2 y^{2} - z^{2}})

\int \frac{cos ( ln ( 9 x ) )}{x} d x

\int (3 x - 2)^{- 11} d x

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

x \to 5 lim (4 x - 3 x^{2})

What is the derivative of f(t)=9tsin(pit) ?
The derivative of f(t)=9tsin(pit) is 9(sin(pit)+pitcos(pit))
What is the first derivative of f(t)=9tsin(pit) ?
The first derivative of f(t)=9tsin(pit) is 9(sin(pit)+pitcos(pit))