Find the derivative of the function.h(t)=sin(t)cos(t)h(t)=\sin\left(t\right)\cos\left(t\right)

− sin ⁡ 2 ( t ) + cos ⁡ 2 ( t ) -\sin^2\left(t\right)+\cos^2\left(t\right) − sin 2 ( t ) + cos 2 ( t )

Find the derivative of the function. h ( t ) = sin ( t ) cos ( t ) h(t)=\sin\left(t\right)\cos\left(t\right)

Here, we're asked to find the derivative of the function h of t equals sine t times cosine t. Now here, I have 2 trig functions multiplying each other, which means that to find the derivative, I need to use the product rule here. So let's go ahead and jump in. Now in order to find my derivative here, we're just taking the first derivative, h prime of t of sine t times cosine t. So using my product rule here, I'm going to take that left function sine of t and then multiply it by the derivative of that right function. Now the derivative of cosine is a negative sine, so I'm gonna be multiplying it by that here. Now that's my first term. Using the product rule, I'm going to add that together with my second term, which is going to be that right function, which is that cosine. So I have cosine of t times the derivative of that left hand function, sine of t. Remember that when using the product rule, we're always going to be taking one original function and then multiplying it by the derivative of the other function. Now here, the derivative of the sine of t is just the positive cosine of t. So this is what I have here for the product rule. But we can do a bit of simplification here because I have sine multiplying negative sine, which means that this is a negative sine squared of t. And then that's multiplying cosine times cosine, which is positive cosine squared of t. Simplifying this as much as possible for my derivative here, I end up getting negative sine squared of t plus cosine squared of t. Now, you could choose to rewrite this as cosine squared t minus sine squared t if you don't like having that negative sign in front. But either of these two answers would be correct. Now, you may be tempted here to simplify this using some trig identities. But because this is a negative sign squared, you can't do that. If it was a positive, then you could, but we don't have to worry about that here. So this is our final answer. Now as you continue to take more and more derivatives dealing with trig functions, you're going to have to refer to all of those derivatives less and less as you commit them to memory. So let's continue getting some more practice dealing with derivatives of trick functions. Thanks for watching, and I'll see you in the next one.

Find the derivative of the function. h ( t ) = sin ⁡ ( t ) cos ⁡ ( t ) h(t)=\sin\left(t\right)\cos\left(t\right)

− sin ⁡ 2 ( t ) + cos ⁡ 2 ( t ) -\sin^2\left(t\right)+\cos^2\left(t\right) − sin 2 ( t ) + cos 2 ( t )

2 sin ⁡ ( t ) cos ⁡ ( t ) 2\sin\left(t\right)\cos\left(t\right) 2 sin ( t ) cos ( t )

3. Techniques of Differentiation

Derivatives of Trig Functions

Calculus

3. Techniques of Differentiation

Derivatives of Trig Functions

Struggling with Calculus?

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Multiple Choice

Find the derivative of the function.
$h (t) = \sin (t) \cos (t)$

$0$

$1$

$2\sin\left(t\right)\cos\left(t\right)$

$-\sin^2\left(t\right)+\cos^2\left(t\right)$

Verified step by step guidance

Identify the function h(t) = sin(t)cos(t). Notice that this is a product of two functions: u(t) = sin(t) and v(t) = cos(t).

To find the derivative of h(t), apply the product rule for differentiation, which states that if h(t) = u(t)v(t), then h'(t) = u'(t)v(t) + u(t)v'(t).

Differentiate u(t) = sin(t) to get u'(t) = cos(t).

Differentiate v(t) = cos(t) to get v'(t) = -sin(t).

Substitute u(t), v(t), u'(t), and v'(t) into the product rule formula: h'(t) = cos(t)cos(t) + sin(t)(-sin(t)). Simplify the expression to get the derivative.

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