What is the correct derivative of sin 5x?

[bobsmith76]

Homework Statement

what is the derivative of sin 5x

The Attempt at a Solution

using the product rule, i think the answer would be

cos 5x + sin 5

but the book says the answer is 5 cos 5x

 

[Scootertaj]

Homework Statement

what is the derivative of sin 5x

The Attempt at a Solution

using the product rule, i think the answer would be

cos 5x + sin 5

but the book says the answer is 5 cos 5x

Take the derivative of sin(x) = cos(x), then take derivative (using power rule) of the inside.
So, derivative of 5x = 5.
Thus, derivative of sin(anything) = cos(anything) * derivative of anything.

 

[cramie10]

Differenciating Sin and Cos is very easy. (It gets trickier with the more complicated trig functions, like Tan). The way I like to remember it, is as follows:

Sin(x)
Cos(x)
-Sin(x)
-Cos(x)

When you differenciate one of these functions, you simply move down the tower to the next function. When you differenciate -Cos(x), you go back up to Sin(x). Except in this case, as you have (5x) inside the brackets, not just (x). So as you're differenciating, you change the Sin(5x) to Cos(5x) and then multiply by the differential of what's inside the brackets. Here, 5x is inside the brackets, the differencial of which is 5. So you multiply by 5. Hence; 5 Cos(5x).

 

[MathWarrior]

[itex]\frac{d}{dx}sin(5x)[/itex]

Chain Rule:
Derivative of inner function times the derivative of the outer function.

[itex]f(g(x)) = f'(g(x))g'(x)[/itex]

Since
[itex]g'(x) = \frac{d}{dx}5x = 5[/itex]

And
[itex]f'(x) = cos(5x)[/itex]

Thus:
[itex]5cos(5x)[/itex]

 

[SammyS]

Homework Statement

what is the derivative of sin 5x

The Attempt at a Solution

using the product rule, i think the answer would be

cos 5x + sin 5

but the book says the answer is 5 cos 5x

You are looking at this as the product of what two quantities ?

 

[bobsmith76]

I see the mistake I was making. As Sammy pointed out I wrongly thought that cos 5x is two quantities when it's not.