What Is The Derivative Of Sin(2X)?

Derivative Of Sin^2X: You would use the chain rule to solve this. To do that, you’ll have to demonstrate what the “outer” gathering is and what the “inner” gathering devised in the outer gathering is.

In this case, $\sin \left(x\right)$ is the inner gathering that is composed as part of the ${\sin }^2\left(x\right)$. To look at it another way, let’s denote =$\sin \left(x\right)$ so that $u^2$=${\sin }^2\left(x\right)$. Do you notice how the composite gathering works here? The outer gathering  of $u^2$ squares the inner congregation of $u=\sin \left(x\right)$. Don’t let the  confuse you, it’s just to show you how one function is a composite of the other. Once you understand this, you can derive.

What Is The Derivative Of Sin^2X

Derivative Of Sin^2(2X)

Differentiate  f(x) = (x³+1)².

The only way we have of doing this so far is by first multiplying out the brackets and then differentiating. If we do this we get

f(x) = x⁶ + 2x³ +1 and therefore f´(x) = 6x⁵ +6x².

This is no problem with a simple example such as the one above but what happens if we have for example f(x) = (x³+1)⁶ ?
In this case it takes far too much effort to heighten out the brackets before differentiating.

To differentiate composite gathering like this we use what is called the Chain Rule. We’ll do example 1 again to see how it works.

f(x) is an example of a composite gathering as was introduced in gathering 2.
It can be composed as f(u) = u² where u = x³+1,

  u is a gathering of x,

that is u(x) = x³+1.

The Chain rule says that we first differentiate f(u) regarding u as the variable and get f´(u) = 2u ( just as (x²)´ = 2x )
Next we differentiate u and get  u´(x) = 3x².  Finally, we multiply the two results together and get 
f´(x) = 2u·3x². Putting back the value of u we get 

f´(x) =2 ( x³+1)·3x² =  6x⁵ +6x²

^{This gives us a rule called the Chain Rule which says that}

Derivative Of Ln(Sin^2X)

To find the derivative of sin(2x), we must understand two basic calculus concepts:

• The derivative of a trigonometric function. These are shown in this table.

• The chain rule, where we take a derivative of a function containing a parenthesis and then multiply that derivative by the derivative of what the parenthesis contains.

Derivative Of Sin^2X)

You would use the chain rule to solve this. To do that, you’ll have to determine what the “outer” function is and what the “inner” function composed in the outer function is.

In this case, $\sin \left(x\right)$ is the inner function that is composed as part of the ${\sin }^2\left(x\right)$. To look at it another way, let’s denote $u$=$\sin \left(x\right)$ so that $u^2$=${\sin }^2\left(x\right)$. Do you notice how the composite function works here? The outer function of $u^2$ squares the inner function of $u=\sin \left(x\right)$. Don’t let the $u$ confuse you, it’s just to show you how one function is a composite of the other. Once you understand this, you can derive.

So, mathematically, the chain rule is:

The derivative of a composite function F(x) is:

F'(x)=f'(g(x))(g'(x))

Or, in words:

the derivative of the outer function (with the inside function left alone!) times the derivative of the inner function.

1) The derivative of the outer function(with the inside function left alone) is:

$\frac{d}{dx}{u}^2=2u$
(I’m leaving the $u$ in for now but you can sub in $u=\sin \left(x\right)$ if you want to while you’re doing the steps. Remember that these are just steps, the actual derivative of the question is shown at the bottom)

2) The derivative of the inner function:

$\frac{d}{dx}\sin \left(x\right)=\cos \left(x\right)$

Combining the two steps through multiplication to get the derivative:

$\frac{d}{dx}{\sin }^2\left(x\right)=2u\cos \left(x\right)=2\sin \left(x\right)\cos \left(x\right)$