How do you differentiate \[{\cos ^4}(x)\]?
Answer
582.9k+ views
Hint: We will use the chain rule of differentiation to differentiate \[{\cos ^4}(x)\]. We will consider \[\cos (x)\] as some variable \[u\]. Then we will find \[{\cos ^4}(x)\] in terms of \[u\] and assign this function to another variable \[v\]. Finally, we will differentiate \[v\] with respect to \[x\] to get the required value.
Formula used:
If \[v = f(u)\] and \[u = f(x)\], then \[\dfrac{{dv}}{{dx}} = \dfrac{{dv}}{{du}} \cdot \dfrac{{du}}{{dx}}\].
Complete step-by-step answer:
We are required to differentiate \[{\cos ^4}(x)\].
We can see that this is a function of \[x\].
Let us differentiate \[{\cos ^4}(x)\] by the method of chain rule of differentiation.
For this, let us take \[\cos (x)\] as some variable \[u\]. So,
\[u = \cos (x)\] ………\[\left( 1 \right)\]
Since \[u\] is a function of \[x\], let us differentiate both sides of the equation \[\left( 1 \right)\] with respect to \[x\]. Therefore, we get
\[ \Rightarrow \dfrac{{du}}{{dx}} = - \sin (x)\] ……..\[\left( 2 \right)\]
But we are required to differentiate \[{\cos ^4}(x)\], where \[u = \cos (x)\]. This means that
\[{u^4} = {\cos ^4}(x)\]
Let u assign this function \[{u^4}\] to a variable \[v\] i.e.,
\[v = {u^4}\] ……….\[\left( 3 \right)\]
We observe that \[v\] is a function of \[u\]. Let us differentiate both sides of the equation \[\left( 3 \right)\] with respect to \[u\].
Now using the formula \[\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}\], where \[x = u\] and \[n = 4\]. This gives us
\[ \Rightarrow \dfrac{{dv}}{{du}} = 4{u^3}\] ……….\[\left( 4 \right)\]
Hence, we observe that \[v\] is a function of \[u\] and in turn \[u\] is a function of \[x\]. So, to differentiate \[v\] with respect to \[x\], we will use the chain rule of differentiation which is as follows:
\[\dfrac{{dv}}{{dx}} = \dfrac{{dv}}{{du}}.\dfrac{{du}}{{dx}}\] ………\[\left( 5 \right)\]
Substituting equations \[\left( 2 \right)\] and \[\left( 4 \right)\] in equation \[\left( 5 \right)\], we get the derivative of \[{\cos ^4}(x)\] as
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\cos }^4}(x)} \right) = (4{u^3}) \cdot \left( { - \sin (x)} \right)\]
Substituting \[u = \cos (x)\] in the above equation, we finally get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\cos }^4}(x)} \right) = - 4{\cos ^3}(x)\sin (x)\]
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 \[{\cos ^4}(x)\], the innermost function is \[\cos (x)\]. This is why we consider this function first, then the function \[{\cos ^4}(x)\].
Formula used:
If \[v = f(u)\] and \[u = f(x)\], then \[\dfrac{{dv}}{{dx}} = \dfrac{{dv}}{{du}} \cdot \dfrac{{du}}{{dx}}\].
Complete step-by-step answer:
We are required to differentiate \[{\cos ^4}(x)\].
We can see that this is a function of \[x\].
Let us differentiate \[{\cos ^4}(x)\] by the method of chain rule of differentiation.
For this, let us take \[\cos (x)\] as some variable \[u\]. So,
\[u = \cos (x)\] ………\[\left( 1 \right)\]
Since \[u\] is a function of \[x\], let us differentiate both sides of the equation \[\left( 1 \right)\] with respect to \[x\]. Therefore, we get
\[ \Rightarrow \dfrac{{du}}{{dx}} = - \sin (x)\] ……..\[\left( 2 \right)\]
But we are required to differentiate \[{\cos ^4}(x)\], where \[u = \cos (x)\]. This means that
\[{u^4} = {\cos ^4}(x)\]
Let u assign this function \[{u^4}\] to a variable \[v\] i.e.,
\[v = {u^4}\] ……….\[\left( 3 \right)\]
We observe that \[v\] is a function of \[u\]. Let us differentiate both sides of the equation \[\left( 3 \right)\] with respect to \[u\].
Now using the formula \[\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}\], where \[x = u\] and \[n = 4\]. This gives us
\[ \Rightarrow \dfrac{{dv}}{{du}} = 4{u^3}\] ……….\[\left( 4 \right)\]
Hence, we observe that \[v\] is a function of \[u\] and in turn \[u\] is a function of \[x\]. So, to differentiate \[v\] with respect to \[x\], we will use the chain rule of differentiation which is as follows:
\[\dfrac{{dv}}{{dx}} = \dfrac{{dv}}{{du}}.\dfrac{{du}}{{dx}}\] ………\[\left( 5 \right)\]
Substituting equations \[\left( 2 \right)\] and \[\left( 4 \right)\] in equation \[\left( 5 \right)\], we get the derivative of \[{\cos ^4}(x)\] as
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\cos }^4}(x)} \right) = (4{u^3}) \cdot \left( { - \sin (x)} \right)\]
Substituting \[u = \cos (x)\] in the above equation, we finally get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\cos }^4}(x)} \right) = - 4{\cos ^3}(x)\sin (x)\]
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 \[{\cos ^4}(x)\], the innermost function is \[\cos (x)\]. This is why we consider this function first, then the function \[{\cos ^4}(x)\].
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