Derivative Of sin^2x, sin^2(2x) & More

Derivative Of sin^2x, sin^2(2x) – The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. Common trigonometric functions include sin(x), cos(x) and tan(x). For example, the derivative of f(x) = sin(x) is represented as f ′(a) = cos(a). f ′(a) is the rate of change of sin(x) at a particular point a.

All derivatives of circular trigonometric functions can be found using those of sin(x) and cos(x). The quotient rule is then implemented to differentiate the resulting expression. Finding the derivatives of the inverse trigonometric functions involves using implicit differentiation and the derivatives of regular trigonometric functions.

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 ${\mathrm{sin²}}^{}\left(x\right)$. To look at it another way, let’s denote $u$=$\sin \left(x\right)$ so that ${\mathrm{u²}}^{}$=${\mathrm{sin²}}^{}\left(x\right)$. Do you notice how the composite function works here? The outer function of ${\mathrm{u²}}^{}$ squares the inner function of $u=\sin \left(x\right)$. Don’t let $u$ confuse you, it’s just to show you how one function is a composite of the other. Once you understand this, you can derive it.

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:

$\frac{\mathrm{d/}}{d\mathrm{x(}}{\mathrm{u²}}^{\mathrm{)}}=2u$
(I’m leaving $u$ 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:

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

Combining the two steps through multiplication to get the derivative:

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

The derivative of sin^2(2x)

Here I would try to use the Chain Rule applied to three functions one nested into the other:
the first and all-embracing function is ${\left(\right)}^2$; the next one is the $\sin$ function and the last one the $2x$ function.

I will use the Chain Rule deriving each one as if alone (regardless of the argument) and I will multiply each individual derivative together using, as visual help, a sequence of red-blue-green colors to identify each derivative:

giving:
$f‘\left(x\right)=4\sin \left(2x\right)\cos \left(2x\right)$

This function can be compressed (giving: $2\sin \left(4x\right)$) using a trigonometric identity but I do not want to confuse the procedure.

what is the derivative of sin^2x

The derivative of ln(sin^2x)

Use the chain rule.

You can break down your function into the logarithm, the square, and the sinus function like follows:

$f\left(u\right)=\ln \left(u\right)$
$u\left(v\right)={\mathrm{v²}}^{}\left(x\right)$
$v\left(x\right)=\sin \left(x\right)$

Now, you need to compute the three derivatives of those three functions (and afterward plug the respective values of $u$ and $v$):

$f‘\left(u\right)=\frac{\mathrm{1/}}{u}=\frac{\mathrm{1/}}{{\mathrm{v²}}^{}}=\frac{\mathrm{1/}}{{\mathrm{sin²}}^{}x}$
$u‘\left(v\right)=2v=2\sin x$
$v‘\left(x\right)=\cos \left(x\right)$

Now, the only thing left to do is multiplying those three derivatives: Hope that this helped!

The derivative of sin^2x + cos^2x

From the given
$y={\mathrm{sin²}}^{}x+{\mathrm{cos²}}^{}x$

The right side of the equation is $=1$

$y=1$

$\frac{d\mathrm{y/}}{dx}=\frac{\mathrm{d/}}{dx}\left(1\right)=0$

or we can do it this way.

God bless I hope the explanation is useful.