What is the derivative of y=sin(9x)+cos(2x) ?

The derivative of y=sin(9x)+cos(2x) is cos(9x)*9-sin(2x)*2

What is the first derivative of y=sin(9x)+cos(2x) ?

The first derivative of y=sin(9x)+cos(2x) is cos(9x)*9-sin(2x)*2

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derivative of

y = sin (9 x) + cos (2 x)

cos (9 x) \cdot 9 - sin (2 x) \cdot 2

Solution steps

\frac{d}{d x} (sin (9 x) + cos (2 x))

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (sin (9 x)) + \frac{d}{d x} (cos (2 x))

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

\frac{d}{d x} (cos (2 x)) = - sin (2 x) \cdot 2

= cos (9 x) \cdot 9 - sin (2 x) \cdot 2

\frac{d}{d x} (3 arcsin (\frac{x}{2}))

\int 2 sin (3 x) d x

n = 0 \sum \infty e^{(- 5 n)}

y^{^{'''}} - y^{^{'}} = sin (x)

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

What is the derivative of y=sin(9x)+cos(2x) ?
The derivative of y=sin(9x)+cos(2x) is cos(9x)*9-sin(2x)*2
What is the first derivative of y=sin(9x)+cos(2x) ?
The first derivative of y=sin(9x)+cos(2x) is cos(9x)*9-sin(2x)*2