What is the derivative of cos(kx+b) ?

The derivative of cos(kx+b) is -sin(kx+b)k

What is the first derivative of cos(kx+b) ?

The first derivative of cos(kx+b) is -sin(kx+b)k

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\frac{d}{d x} (cos (k x + b))

- sin (k x + b) k

Solution steps

\frac{d}{d x} (cos (k x + b))

Apply the chain rule:

- sin (k x + b) \frac{d}{d x} (k x + b)

= - sin (k x + b) \frac{d}{d x} (k x + b)

\frac{d}{d x} (k x + b) = k

= - sin (k x + b) k

y + x y^{^{'}} = 2 e^{2 x}

\frac{d}{d x} (\frac{1}{4} sin^{2} (2 x))

\frac{d y}{d x} = \frac{( y ^{2} + 2 x y )}{x ^{2}}

n = 1 \sum \infty 3 e^{- 5 n}

f (x) = 9 + 4 x^{2} - 2 x^{3}, at x = 2

What is the derivative of cos(kx+b) ?
The derivative of cos(kx+b) is -sin(kx+b)k
What is the first derivative of cos(kx+b) ?
The first derivative of cos(kx+b) is -sin(kx+b)k