Differentiate #1/sinx^2# using chain rule?

Question

Elijah Abram · Accepted Answer

#(df)/(dx)=-2xcotx^2csc^2x^2#

We use chain rule here.

Using this in order to differentiate a function of a function, say #y, =f(g(x))#, where we have to find #(dy)/(dx)#, we need to do (a) substitute #u=g(x)#, which gives us #y=f(u)#. Then we need to use a formula called Chain Rule, which states that #(dy)/(dx)=(dy)/(du)xx(du)/(dx)#.

In fact if we have something like #y=f(g(h(x)))#, we can have #(dy)/(dx)=(dy)/(df)xx(df)/(dg)xx(dg)/(dh)#

Here we have #f(x)=1/(g(x))#, where #g(x)=sin(h(x))# and #h(x)=x^2#.

Hence, #(df)/(dx)=-1/(sin^2x^2)xxcosx^2xx2x=-2xcotx^2csc^2x^2#

Christopher Allers · Answer

#-2xcos(x^2)/sin^2(x^2)# We can use the chain rule to differentiate the function, the chain rule states #d/dx f(g(x)) = f'(g(x)) * g'(x)# So #d/dx (sin(x^2))^-1 = -1sin^-2(x^2)*cos(x^2)*2x = -2xcos(x^2)/sin^2(x^2)#

Ezra Adams · Answer

undefined To differentiate $ \frac{1}{{\sin^2(x)}} $ using the chain rule, follow these steps:

1. Recognize that the function is composed of two functions: the reciprocal function $ \frac{1}{u} $ and the function $ u = \sin(x) $.
2. Apply the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.
3. Let $ u = \sin(x) $, then $ \frac{d}{dx}(\sin(x)) = \cos(x) $.
4. Now, apply the chain rule:

$$
\frac{d}{dx}\left(\frac{1}{{\sin^2(x)}}\right) = \frac{d}{du}\left(\frac{1}{u^2}\right) \cdot \frac{d}{dx}(\sin(x)) = -\frac{2}{u^3} \cdot \cos(x)
$$

5. Substitute $ u = \sin(x) $ back in:

$$
\frac{d}{dx}\left(\frac{1}{{\sin^2(x)}}\right) = -\frac{2}{{\sin^3(x)}} \cdot \cos(x)
$$

Therefore, the derivative of $ \frac{1}{{\sin^2(x)}} $ with respect to $ x $ using the chain rule is $ -\frac{2}{{\sin^3(x)}} \cdot \cos(x) $.