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Popular >

d/(dt)(atcos(4t)+btsin(4t))

Solution

\frac{d}{d t} (a t cos (4 t) + b t sin (4 t))

Solution

a (cos (4 t) - 4 t sin (4 t)) + b (sin (4 t) + 4 t cos (4 t))

Solution steps

\frac{d}{d t} (a t cos (4 t) + b t sin (4 t))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (a t cos (4 t)) + \frac{d}{d t} (b t sin (4 t))

\frac{d}{d t} (a t cos (4 t)) = a (cos (4 t) - 4 t sin (4 t))

\frac{d}{d t} (b t sin (4 t)) = b (sin (4 t) + 4 t cos (4 t))

= a (cos (4 t) - 4 t sin (4 t)) + b (sin (4 t) + 4 t cos (4 t))

Popular Examples

(\partial)/(\partial x)(-x^2sin(x^2y))

\frac{\partial}{\partial x} (- x^{2} sin (x^{2} y))

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n \to \infty lim (\frac{4}{n})

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\frac{d}{d x} ((2 x + e^{x}) (5 - x))

partialfraction (2x^2)/(x^2-1)partial fractions

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(\partial)/(\partial x)(e^{x^2y^4}sin(xyz))

\frac{\partial}{\partial x} (e^{x^{2} y^{4}} sin (x yz))

Frequently Asked Questions (FAQ)

What is the d/(dt)(atcos(4t)+btsin(4t)) ?
The d/(dt)(atcos(4t)+btsin(4t)) is a(cos(4t)-4tsin(4t))+b(sin(4t)+4tcos(4t))
What is the first d/(dt)(atcos(4t)+btsin(4t)) ?
The first d/(dt)(atcos(4t)+btsin(4t)) is a(cos(4t)-4tsin(4t))+b(sin(4t)+4tcos(4t))