What is the derivative of y=19cos(x/b-2.5)+19 ?

The derivative of y=19cos(x/b-2.5)+19 is (19(0.80114…sin(x/b)+0.59847…cos(x/b)))/b

What is the first derivative of y=19cos(x/b-2.5)+19 ?

The first derivative of y=19cos(x/b-2.5)+19 is (19(0.80114…sin(x/b)+0.59847…cos(x/b)))/b

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derivative of

y = 19 cos (\frac{x}{b} - 2.5) + 19

\frac{19 ( 0.80114 \dots sin ( \frac{x}{b} ) + 0.59847 \dots cos ( \frac{x}{b} ) )}{b}

Solution steps

\frac{d}{d x} (19 cos (\frac{x}{b} - 2.5) + 19)

Simplify

19 cos (\frac{x}{b} - 2.5) + 19 : 19 (- 0.80114 \dots cos (\frac{x}{b}) + 0.59847 \dots sin (\frac{x}{b})) + 19

= \frac{d}{d x} (19 (- 0.80114 \dots cos (\frac{x}{b}) + 0.59847 \dots sin (\frac{x}{b})) + 19)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (19 (- 0.80114 \dots cos (\frac{x}{b}) + 0.59847 \dots sin (\frac{x}{b}))) + \frac{d}{d x} (19)

\frac{d}{d x} (19 (- 0.80114 \dots cos (\frac{x}{b}) + 0.59847 \dots sin (\frac{x}{b}))) = \frac{19 ( 0.80114 \dots sin ( \frac{x}{b} ) + 0.59847 \dots cos ( \frac{x}{b} ) )}{b}

\frac{d}{d x} (19) = 0

= \frac{19 ( 0.80114 \dots sin ( \frac{x}{b} ) + 0.59847 \dots cos ( \frac{x}{b} ) )}{b} + 0

Simplify

= \frac{19 ( 0.80114 \dots sin ( \frac{x}{b} ) + 0.59847 \dots cos ( \frac{x}{b} ) )}{b}

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\frac{\partial}{\partial x} (\frac{t}{x})

\frac{d y}{d x} + (\frac{3}{x}) y = - 4 x - 10

\int \frac{1}{x ^{2} - 4} d x

What is the derivative of y=19cos(x/b-2.5)+19 ?
The derivative of y=19cos(x/b-2.5)+19 is (19(0.80114…sin(x/b)+0.59847…cos(x/b)))/b
What is the first derivative of y=19cos(x/b-2.5)+19 ?
The first derivative of y=19cos(x/b-2.5)+19 is (19(0.80114…sin(x/b)+0.59847…cos(x/b)))/b