What is the derivative of y=cos((pit)/4-pi/3) ?

The derivative of y=cos((pit)/4-pi/3) is -pi/8 sin((pit)/4)+(sqrt(3)pi)/8 cos((pit)/4)

What is the first derivative of y=cos((pit)/4-pi/3) ?

The first derivative of y=cos((pit)/4-pi/3) is -pi/8 sin((pit)/4)+(sqrt(3)pi)/8 cos((pit)/4)

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derivative of

y = cos (\frac{π t}{4} - \frac{π}{3})

- \frac{π}{8} sin (\frac{π t}{4}) + \frac{3 π}{8} cos (\frac{π t}{4})

Solution steps

\frac{d}{d t} (cos (\frac{π t}{4} - \frac{π}{3}))

Simplify

cos (\frac{π t}{4} - \frac{π}{3}) : \frac{1}{2} cos (\frac{π t}{4}) + \frac{3}{2} sin (\frac{π t}{4})

= \frac{d}{d t} (\frac{1}{2} cos (\frac{π t}{4}) + \frac{3}{2} sin (\frac{π t}{4}))

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (\frac{1}{2} cos (\frac{π t}{4})) + \frac{d}{d t} (\frac{3}{2} sin (\frac{π t}{4}))

\frac{d}{d t} (\frac{1}{2} cos (\frac{π t}{4})) = - \frac{π}{8} sin (\frac{π t}{4})

\frac{d}{d t} (\frac{3}{2} sin (\frac{π t}{4})) = \frac{3 π}{8} cos (\frac{π t}{4})

= - \frac{π}{8} sin (\frac{π t}{4}) + \frac{3 π}{8} cos (\frac{π t}{4})

\int \frac{2}{x ^{5}} d x

\int cot (u) d u

y^{^{'}} = 2 x (x - 2)

n = 0 \sum \infty \frac{1 ^{n}}{n + 2}

\int_{0}^{5} \frac{1}{( 6 - 2 x ) ^{\frac{1}{3}}} d x

What is the derivative of y=cos((pit)/4-pi/3) ?
The derivative of y=cos((pit)/4-pi/3) is -pi/8 sin((pit)/4)+(sqrt(3)pi)/8 cos((pit)/4)
What is the first derivative of y=cos((pit)/4-pi/3) ?
The first derivative of y=cos((pit)/4-pi/3) is -pi/8 sin((pit)/4)+(sqrt(3)pi)/8 cos((pit)/4)