Solveeit Logo
Login

Question

Question: How do you differentiate \(f\left( x \right) = \dfrac{{\cos x}}{{\sin x}}\) ?...

How do you differentiate f(x)=cos⁡xsin⁡xf\left( x \right) = \dfrac{{\cos x}}{{\sin x}} ?

Explanation

Solution

In this question, we have been given a trigonometric equation and we have been asked to differentiate the equation. This question can be done in two ways.
Method 1: Use quotient rule to find the differentiation. Once you have put the values in the quotient rule, you will see that a trigonometric formula will be used here. Put that formula in the equation and you will have your answer.
Method 2: Use trigonometry to simplify the given ratio. You will get a trigonometric ratio. Simply, write its differentiation and you will get the answer. But, if you are bad at remembering formulas, refer to the method 1 above.

Formula used: 1) Quotient rule: ddx[f(x)g(x)]=g(x).f′(x)−f(x)g′(x)(g(x))2\dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right).f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}
2) d(cot⁡x)dx=cos⁡ec2x\dfrac{{d\left( {\cot x} \right)}}{{dx}} = \cos e{c^2}x

Complete step-by-step solution:
We are given a trigonometric equation and we have to find its differentiation. I will use method 1 here.
In this method, we will simply use quotient rules. The quotient rule says that –
ddx[f(x)g(x)]=g(x)f′(x)−f(x)g′(x)(g(x))2\dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}
⇒f(x)=cos⁡xsin⁡x\Rightarrow f\left( x \right) = \dfrac{{\cos x}}{{\sin x}} …. (given)
Differentiating both the sides with respect to x by applying the quotient rule,
⇒f′(x)=cos⁡x(sin⁡x)′−sin⁡x(cos⁡x)′sin⁡2x\Rightarrow f'\left( x \right) = \dfrac{{\cos x\left( {\sin x} \right)' - \sin x\left( {\cos x} \right)'}}{{{{\sin }^2}x}}
Simplifying the equation,
⇒f′(x)=cos⁡xcos⁡x+sin⁡xsin⁡xsin⁡2x\Rightarrow f'\left( x \right) = \dfrac{{\cos x\cos x + \sin x\sin x}}{{{{\sin }^2}x}}
⇒f′(x)=cos⁡2x+sin⁡2xsin⁡2x\Rightarrow f'\left( x \right) = \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\sin }^2}x}}
Now, we know that cos⁡2x+sin⁡2x=1{\cos ^2}x + {\sin ^2}x = 1. Using this in the equation,
⇒f′(x)=1sin⁡2x\Rightarrow f'\left( x \right) = \dfrac{1}{{{{\sin }^2}x}}
We also know that 1sin⁡x=cos⁡ecx\dfrac{1}{{\sin x}} = \cos ecx. Using this to simplify the equation,
⇒f′(x)=cos⁡ec2x\Rightarrow f'\left( x \right) = \cos e{c^2}x

Hence, the differentiation of cos⁡xsin⁡x\dfrac{{\cos x}}{{\sin x}} is cos⁡ec2x\cos e{c^2}x.

Note: Now, I will show you how we can solve the given equation using method 2.
⇒f(x)=cos⁡xsin⁡x\Rightarrow f\left( x \right) = \dfrac{{\cos x}}{{\sin x}} …. (given)
We know that cos⁡xsin⁡x=cot⁡x\dfrac{{\cos x}}{{\sin x}} = \cot x. Using this, we get,
⇒f(x)=cot⁡x\Rightarrow f\left( x \right) = \cot x
Differentiating both the sides with respect to x,
f′(x)=cos⁡ec2xf'\left( x \right) = \cos e{c^2}x
Hence, this is our final answer and it is similar to the answer we got using the previous method.