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The derivative of y = sin(3x + 5) is 3cos(3x + 5) but only if x is measured in radians.

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What is the derivative for 3x 1?

If y = 3x +- 1, the derivative with respect to x is y' = 3.


Find the derivatives of y equals 2 sin 3x and show the solution?

y = 2 sin 3x y' = 2(sin 3x)'(3x)' y' = 2(cos 3x)(3) y' = 6 cos 3x


What is the amplitude of the function y -3 sin 3x?

amplitude of the function y =-3 sin 3x


The derivative of y x2 - 3x at the point?

-1


What is the domain of the function y sin x?

y= sin 3x


Find the derivative y equals 10 to the sin3x power?

Y=10^sin(x) The derivative is: (log(5)+log(2))*cos(x)*2^sin(x)*5^sin(x) Use the chain rule, product rule, and power rules combined with sin(x) rule.


Find the derivatives of y equals 2 sin 3x?

y=2 sin(3x) dy/dx = 2 cos(3x) (3) dy/dx = 6 cos(3x)


How do you differentiate sin sin x?

To differentiate y=sin(sin(x)) you need to use the chain rule. A common way to remember the chain rule is "derivative of the outside, keep the inside, derivative of the inside". First, you take the derivative of the outside. The derivative of sin is cos. Then, you keep the inside, so you keep sin(x). Then, you multiple by the derivative of the inside. Again, the derivative of sinx is cosx. In the end, you get y'=cos(sin(x))cos(x))


What is the derivative of y equals sin times x to the power of 2?

D(y)= sin 2x


3x - y less than or equal to 5?

-2


What is the derivative of sinx pwr cosx?

For the function: y = sin(x)cos(x) To find the derivative y', implicit differentiation must be used. To do this, both sides of the equation must be put into the argument of a natural logarithm: ln(y) = ln(sin(x)cos(x)) by the properties of logarithms, this can also be expressed as: ln(y) = cos(x)ln(sin(x)) deriving both sides of the equation yields: (1/y)(y') = cos(x)(1/sin(x))(cos(x)) + -sin(x)ln(sin(x)) This derivative features two important things. The obvious thing is the product rule use to differentiate the right side of the equation. The left side of the equation brings into play the "implicit" differentiation part of this problem. The derivative of ln(y) is a chain rule. The derivative of just ln(y) is simply 1/y, but you must also multiply by the derivative of y, which is y'. so the total derivative of ln(y) is (1/y)(y'). solving for y' in the above, the following is found: y' = y[(cos2(x)/sin(x)) - sin(x)ln(sin(x))] = y[cot(x)cos(x) - sin(x)ln(sin(x))] y' = y[cot(x)cos(x) - sin(x)ln(sin(x))] = sin(x)cos(x)[cot(x)cos(x) - sin(x)ln(sin(x)) is the most succinct form of this derivative.


What is the rate of change in the function y equals tan3x?

In this case, you'll need to apply the chain rule, first taking the derivative of the tan function, and multiplying by the derivative of 3x: y = tan(3x) ∴ dy/dx = 3sec2(3x)

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