What is the derivative of \[\cos {x^2}\] \[?\]
Answer
557.7k+ views
Hint: The composite function is a function where one function is substituting into another function. Since we know that \[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\] .
Suppose \[u\] and \[v\] are functions of \[x\] only. Then differentiation has the following rules:
1. The derivative of the constant function is equal to zero.
2. Product rule: \[\dfrac{d}{{dx}}\left( {u \times v} \right) = v \times \dfrac{d}{{dx}}\left( u \right) + u \times \dfrac{d}{{dx}}\left( v \right)\] .
3. Quotient rule: \[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v \times \dfrac{d}{{dx}}\left( u \right) - u \times \dfrac{d}{{dx}}\left( v \right)}}{{{v^2}}}\] .
4. If \[u\] is a function of \[v\], Then \[\dfrac{d}{{dx}}\left( {u(v)} \right) = \dfrac{d}{{dv}}\left( {u(v)} \right).\dfrac{{dv}}{{dx}}\]
Complete step by step solution:
Given \[\cos {x^2}\] ----(1)
Note that \[{4^{th}}\] rule is known as the chain rule. If the function is composed of three functions, say \[u\], \[v\] and \[w\] are functions of \[x\] . Then the derivative of the composition of three function is defined as follows
\[\dfrac{d}{{dx}}\left( {u(v(w(x)))} \right) = \dfrac{{du}}{{dv}}.\dfrac{{dv}}{{dw}}.\dfrac{{dw}}{{dx}}\] .
Differentiating with respect to \[x\] both sides of the equation (1), we get
\[\dfrac{d}{{dx}}\left( {\cos {x^2}} \right) = - \sin {x^2}\dfrac{d}{{dx}}\left( {{x^2}} \right) = - 2x\sin {x^2}\] .
Hence, the derivative of the given function \[\cos {x^2}\] is \[ - 2x\sin {x^2}\]
So, the correct answer is “ \[ - 2x\sin {x^2}\] ”.
Note: A differential equation is the equation which contains dependent variables, independent variables and derivatives of the dependent variables with respect to the independent variables. Since differential equations are classified into two types, Ordinary differential equations where dependent variables depend on only one independent variable and Partial differential equations where dependent variables depend on two or more independent variables.
Suppose \[u\] and \[v\] are functions of \[x\] only. Then differentiation has the following rules:
1. The derivative of the constant function is equal to zero.
2. Product rule: \[\dfrac{d}{{dx}}\left( {u \times v} \right) = v \times \dfrac{d}{{dx}}\left( u \right) + u \times \dfrac{d}{{dx}}\left( v \right)\] .
3. Quotient rule: \[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v \times \dfrac{d}{{dx}}\left( u \right) - u \times \dfrac{d}{{dx}}\left( v \right)}}{{{v^2}}}\] .
4. If \[u\] is a function of \[v\], Then \[\dfrac{d}{{dx}}\left( {u(v)} \right) = \dfrac{d}{{dv}}\left( {u(v)} \right).\dfrac{{dv}}{{dx}}\]
Complete step by step solution:
Given \[\cos {x^2}\] ----(1)
Note that \[{4^{th}}\] rule is known as the chain rule. If the function is composed of three functions, say \[u\], \[v\] and \[w\] are functions of \[x\] . Then the derivative of the composition of three function is defined as follows
\[\dfrac{d}{{dx}}\left( {u(v(w(x)))} \right) = \dfrac{{du}}{{dv}}.\dfrac{{dv}}{{dw}}.\dfrac{{dw}}{{dx}}\] .
Differentiating with respect to \[x\] both sides of the equation (1), we get
\[\dfrac{d}{{dx}}\left( {\cos {x^2}} \right) = - \sin {x^2}\dfrac{d}{{dx}}\left( {{x^2}} \right) = - 2x\sin {x^2}\] .
Hence, the derivative of the given function \[\cos {x^2}\] is \[ - 2x\sin {x^2}\]
So, the correct answer is “ \[ - 2x\sin {x^2}\] ”.
Note: A differential equation is the equation which contains dependent variables, independent variables and derivatives of the dependent variables with respect to the independent variables. Since differential equations are classified into two types, Ordinary differential equations where dependent variables depend on only one independent variable and Partial differential equations where dependent variables depend on two or more independent variables.
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