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Question

Question: How do you differentiate \[{\cos ^4}(x)\]?...

How do you differentiate cos4(x){\cos ^4}(x)?

Explanation

Solution

We will use the chain rule of differentiation to differentiate cos4(x){\cos ^4}(x). We will consider cos(x)\cos (x) as some variable uu. Then we will find cos4(x){\cos ^4}(x) in terms of uu and assign this function to another variable vv. Finally, we will differentiate vv with respect to xx to get the required value.

Formula used:
If v=f(u)v = f(u) and u=f(x)u = f(x), then dvdx=dvdududx\dfrac{{dv}}{{dx}} = \dfrac{{dv}}{{du}} \cdot \dfrac{{du}}{{dx}}.

Complete step-by-step answer:
We are required to differentiate cos4(x){\cos ^4}(x).
We can see that this is a function of xx.
Let us differentiate cos4(x){\cos ^4}(x) by the method of chain rule of differentiation.
For this, let us take cos(x)\cos (x) as some variable uu. So,
u=cos(x)u = \cos (x) ………(1)\left( 1 \right)
Since uu is a function of xx, let us differentiate both sides of the equation (1)\left( 1 \right) with respect to xx. Therefore, we get
dudx=sin(x)\Rightarrow \dfrac{{du}}{{dx}} = - \sin (x) ……..(2)\left( 2 \right)
But we are required to differentiate cos4(x){\cos ^4}(x), where u=cos(x)u = \cos (x). This means that
u4=cos4(x){u^4} = {\cos ^4}(x)
Let u assign this function u4{u^4} to a variable vv i.e.,
v=u4v = {u^4} ……….(3)\left( 3 \right)
We observe that vv is a function of uu. Let us differentiate both sides of the equation (3)\left( 3 \right) with respect to uu.
Now using the formula ddx(xn)=nxn1\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}, where x=ux = u and n=4n = 4. This gives us
dvdu=4u3\Rightarrow \dfrac{{dv}}{{du}} = 4{u^3} ……….(4)\left( 4 \right)
Hence, we observe that vv is a function of uu and in turn uu is a function of xx. So, to differentiate vv with respect to xx, we will use the chain rule of differentiation which is as follows:
dvdx=dvdu.dudx\dfrac{{dv}}{{dx}} = \dfrac{{dv}}{{du}}.\dfrac{{du}}{{dx}} ………(5)\left( 5 \right)
Substituting equations (2)\left( 2 \right) and (4)\left( 4 \right) in equation (5)\left( 5 \right), we get the derivative of cos4(x){\cos ^4}(x) as
ddx(cos4(x))=(4u3)(sin(x))\Rightarrow \dfrac{d}{{dx}}\left( {{{\cos }^4}(x)} \right) = (4{u^3}) \cdot \left( { - \sin (x)} \right)
Substituting u=cos(x)u = \cos (x) in the above equation, we finally get
ddx(cos4(x))=4cos3(x)sin(x)\Rightarrow \dfrac{d}{{dx}}\left( {{{\cos }^4}(x)} \right) = - 4{\cos ^3}(x)\sin (x)

Note: The chain rule of differentiation allows the differentiation of composite functions. To use this rule, we must start substituting functions from the innermost function. Then, we must work our way outwards, and at each step, we must find the derivative. In cos4(x){\cos ^4}(x), the innermost function is cos(x)\cos (x). This is why we consider this function first, then the function cos4(x){\cos ^4}(x).