What is the derivative of 9sin(x+9cos(x)) ?

The derivative of 9sin(x+9cos(x)) is 9cos(x)-9sin(x)

What is the first derivative of 9sin(x+9cos(x)) ?

The first derivative of 9sin(x+9cos(x)) is 9cos(x)-9sin(x)

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\frac{d}{d x} (9 sin (x) + 9 cos (x))

9 cos (x) - 9 sin (x)

Solution steps

\frac{d}{d x} (9 sin (x) + 9 cos (x))

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (9 sin (x)) + \frac{d}{d x} (9 cos (x))

\frac{d}{d x} (9 sin (x)) = 9 cos (x)

\frac{d}{d x} (9 cos (x)) = - 9 sin (x)

= 9 cos (x) - 9 sin (x)

x^{\frac{x}{3}}

f (x) = e^{2 x}, at x = \frac{1}{2} ln (2)

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

f^{^{'}} (x) = 6 x^{2}

f (2) = 5 - 2 x

What is the derivative of 9sin(x+9cos(x)) ?
The derivative of 9sin(x+9cos(x)) is 9cos(x)-9sin(x)
What is the first derivative of 9sin(x+9cos(x)) ?
The first derivative of 9sin(x+9cos(x)) is 9cos(x)-9sin(x)