What is the derivative of 1.2*sin(2*x-0.8*t+10) ?

The derivative of 1.2*sin(2*x-0.8*t+10) is 2.4cos(2x-0.8t+10)

What is the first derivative of 1.2*sin(2*x-0.8*t+10) ?

The first derivative of 1.2*sin(2*x-0.8*t+10) is 2.4cos(2x-0.8t+10)

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\frac{d}{d x} (1.2 \cdot sin (2 \cdot x - 0.8 \cdot t + 10))

2.4 cos (2 x - 0.8 t + 10)

Solution steps

\frac{d}{d x} (1.2 sin (2 x - 0.8 t + 10))

Treat

t

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 1.2 \frac{d}{d x} (sin (2 x - 0.8 t + 10))

Apply the chain rule:

cos (2 x - 0.8 t + 10) \frac{d}{d x} (2 x - 0.8 t + 10)

= cos (2 x - 0.8 t + 10) \frac{d}{d x} (2 x - 0.8 t + 10)

\frac{d}{d x} (2 x - 0.8 t + 10) = 2

= 1.2 cos (2 x - 0.8 t + 10) \cdot 2

Simplify

= 2.4 cos (2 x - 0.8 t + 10)

\frac{ln ( x )}{2}

3 sec^{2} (x)

cot^{3} (x)

sin (x + 2)

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

What is the derivative of 1.2*sin(2*x-0.8*t+10) ?
The derivative of 1.2*sin(2*x-0.8*t+10) is 2.4cos(2x-0.8t+10)
What is the first derivative of 1.2*sin(2*x-0.8*t+10) ?
The first derivative of 1.2*sin(2*x-0.8*t+10) is 2.4cos(2x-0.8t+10)