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Question: If \[x = {\cos ^2}\theta \] and \[y = \cot \theta \], then find \[\dfrac{{dy}}{{dx}}\]at \[\theta = ...

If x=cos2θx = {\cos ^2}\theta and y=cotθy = \cot \theta , then find dydx\dfrac{{dy}}{{dx}}at θ=π4\theta = \dfrac{\pi }{4}

Explanation

Solution

Here, we will use the concept of differentiation to find the value of dydx\dfrac{{dy}}{{dx}}. We will first find the value of the dxdx and then we will find the value of the dydy separately. Then by dividing their values we will get the value of dydx\dfrac{{dy}}{{dx}}. Further, we will substitute the value of θ=π4\theta = \dfrac{\pi }{4} to get the final answer.

Complete step-by-step answer:
We will first find the value of dxdx by differentiating the equation ,x=cos2θx = {\cos ^2}\theta .
We know that the differentiation of cosθ\cos \theta is sinθ- \sin \theta.
Differentiating the equation x=cos2θx = {\cos ^2}\theta , we get
dx=2cosθsinθdθdx = - 2\cos \theta \sin \theta \,d\theta ……………….(1)\left( 1 \right)
Now we will find the value of dydy by differentiating the equation y=cotθy = \cot \theta .
We know that the differentiation of cotθ\cot \theta is cosec2θ- \cos e{c^2}\theta. So, we get
dy=cosec2θdθdy = - \cos e{c^2}\theta \,d\theta ……………….(2)\left( 2 \right)
Now by dividing the equation (2)\left( 2 \right) by (1)\left( 1 \right), we get
dydx=cosec2θdθ2cosθsinθdθ\dfrac{{dy}}{{dx}} = \dfrac{{ - \cos e{c^2}\theta \,d\theta }}{{ - 2\cos \theta \sin \theta \,d\theta }}
We know that cosecθ{\rm{cosec}}\theta is the reciprocal of the sinθ\sin \theta . Therefore, we get
dydx=12sin2θcosθsinθ\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2{{\sin }^2}\theta \cos \theta \sin \theta }}
Now we will find the value of the dydx\dfrac{{dy}}{{dx}} at θ=π4\theta = \dfrac{\pi }{4}.
We know that the value of sinπ4=12\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} and cosπ4=12\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}.
Substituting sinπ4=12\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} and cosπ4=12\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} in the equation, we get
dydx=12×(12)2×12×12\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2 \times {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2} \times \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }}}}
Simplifying the equation, we get
dydx=12×12×12=2\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2 \times \dfrac{1}{2} \times \dfrac{1}{2}}} = 2
Hence, dydx\dfrac{{dy}}{{dx}} at θ=π4\theta = \dfrac{\pi }{4} is 2.

Note: Here, we need to know the basic differentiation of the trigonometric function in order to solve questions. We have used differentiation by parts to find the value of dxdx. Differentiation is a method by which we can measure per unit of a function in the given independent variable.