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Popular >

d/(dt)(-2asin(2t)+bcos(2t)*2)

Solution

\frac{d}{d t} (- 2 a sin (2 t) + b cos (2 t) \cdot 2)

Solution

- 4 a cos (2 t) - 4 b sin (2 t)

Solution steps

\frac{d}{d t} (- 2 a sin (2 t) + b cos (2 t) \cdot 2)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d t} (2 a sin (2 t)) + \frac{d}{d t} (b cos (2 t) \cdot 2)

\frac{d}{d t} (2 a sin (2 t)) = 4 a cos (2 t)

\frac{d}{d t} (b cos (2 t) \cdot 2) = - 4 b sin (2 t)

= - 4 a cos (2 t) - 4 b sin (2 t)

Popular Examples

integral of sqrt((x^2-2)/(x^8))

\int \frac{x ^{2} - 2}{x ^{8}} d x

integral of (cos^5(α))/(sqrt(sin(α)))

\int \frac{cos ^{5} ( α )}{sin ( α )} d α

integral of cos(x)(4+7sin^2(x))

\int cos (x) (4 + 7 sin^{2} (x)) d x

(\partial)/(\partial y)(3xy^2z^2)

\frac{\partial}{\partial y} (3 x y^{2} z^{2})

f^'(t)= 6/(t^6)-3/(t^3)+2/t

f^{^{'}} (t) = \frac{6}{t ^{6}} - \frac{3}{t ^{3}} + \frac{2}{t}

Frequently Asked Questions (FAQ)

What is the d/(dt)(-2asin(2t)+bcos(2t)*2) ?
The d/(dt)(-2asin(2t)+bcos(2t)*2) is -4acos(2t)-4bsin(2t)
What is the first d/(dt)(-2asin(2t)+bcos(2t)*2) ?
The first d/(dt)(-2asin(2t)+bcos(2t)*2) is -4acos(2t)-4bsin(2t)