What is the derivative of-asin(x+bcos(x)) ?

The derivative of-asin(x+bcos(x)) is -acos(x)-bsin(x)

What is the first derivative of-asin(x+bcos(x)) ?

The first derivative of-asin(x+bcos(x)) is -acos(x)-bsin(x)

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

- a cos (x) - b sin (x)

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (a sin (x)) = a cos (x)

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

= - a cos (x) - b sin (x)

f (x) = 4 x

\frac{\partial}{\partial x} (- 15 (3 x + y)^{4})

y^{^{'}} = (y - 1)^{2}

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

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

What is the derivative of-asin(x+bcos(x)) ?
The derivative of-asin(x+bcos(x)) is -acos(x)-bsin(x)
What is the first derivative of-asin(x+bcos(x)) ?
The first derivative of-asin(x+bcos(x)) is -acos(x)-bsin(x)