What is the derivative of y=55cos(0.071x+0.39)+63 ?

The derivative of y=55cos(0.071x+0.39)+63 is 55(-0.06566…sin(0.071x)-0.02699…cos(0.071x))

What is the first derivative of y=55cos(0.071x+0.39)+63 ?

The first derivative of y=55cos(0.071x+0.39)+63 is 55(-0.06566…sin(0.071x)-0.02699…cos(0.071x))

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derivative of

y = 55 cos (0.071 x + 0.39) + 63

55 (- 0.06566 \dots sin (0.071 x) - 0.02699 \dots cos (0.071 x))

Solution steps

\frac{d}{d x} (55 cos (0.071 x + 0.39) + 63)

Simplify

55 cos (0.071 x + 0.39) + 63 : 55 (0.92490 \dots cos (0.071 x) - 0.38018 \dots sin (0.071 x)) + 63

= \frac{d}{d x} (55 (0.92490 \dots cos (0.071 x) - 0.38018 \dots sin (0.071 x)) + 63)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (55 (0.92490 \dots cos (0.071 x) - 0.38018 \dots sin (0.071 x))) + \frac{d}{d x} (63)

\frac{d}{d x} (55 (0.92490 \dots cos (0.071 x) - 0.38018 \dots sin (0.071 x))) = 55 (- 0.06566 \dots sin (0.071 x) - 0.02699 \dots cos (0.071 x))

\frac{d}{d x} (63) = 0

= 55 (- 0.06566 \dots sin (0.071 x) - 0.02699 \dots cos (0.071 x)) + 0

Simplify

= 55 (- 0.06566 \dots sin (0.071 x) - 0.02699 \dots cos (0.071 x))

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\frac{d v}{d t} = v^{2} + 5 v + 2

\frac{d}{d x} (\frac{4}{3} x^{3} + \frac{1}{2} x - 4)

What is the derivative of y=55cos(0.071x+0.39)+63 ?
The derivative of y=55cos(0.071x+0.39)+63 is 55(-0.06566…sin(0.071x)-0.02699…cos(0.071x))
What is the first derivative of y=55cos(0.071x+0.39)+63 ?
The first derivative of y=55cos(0.071x+0.39)+63 is 55(-0.06566…sin(0.071x)-0.02699…cos(0.071x))