How do you differentiate $f(x)=\csc x$ using the quotient rule?
Answer
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Hint: We are given an expression which we have to differentiate using the quotient rule. First, we will write the expression in simplified form and we get, \[f(x)=\dfrac{1}{\sin x}\]. We will then use the quotient rule and substitute the values in the formula as required. On solving further, we get the differentiation of the given expression.
Complete step-by-step solution:
According to the given question, we are given an expression which we have to differentiate using the
quotient rule.
Quotient rule is a rule for differentiation of function which involves ratio of two functions which are
differentiable. The formula of the quotient rule is as follows.
\[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}(u)-u\dfrac{d}{dx}(v)}{{{v}^{2}}}\]
We will carry out the differentiation based on the above formula of quotient rule.
The given expression we have is,
$f(x)=\csc x$-----(1)
We can simplify it further and write the equation (1) as,
\[f(x)=\dfrac{1}{\sin x}\]-----(2)
Using the formula of quotient rule, we will substitute the values in it. And we have,
\[f'(x)=\dfrac{d}{dx}\left( \dfrac{1}{\sin x} \right)=\dfrac{\sin x\dfrac{d}{dx}(1)-1\dfrac{d}{dx}(\sin
x)}{{{\sin }^{2}}x}\]-----(3)
We proceed on to solve the above expression to get the differentiation value, we have,
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{\sin x} \right)=\dfrac{\sin x(0)-1(\cos x)}{{{\sin }^{2}}x}\]
And as we know that, differentiation of any constant gives us a value 0, that is, \[\dfrac{d}{dx}(1)=0\]
and differentiation of sine function is cosine function, that is, \[\dfrac{d}{dx}(\sin x)=\cos x\].
So, we get,
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{\sin x} \right)=\dfrac{-(\cos x)}{{{\sin }^{2}}x}\]
We can further proceed as,
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{\sin x} \right)=\dfrac{-\cos x}{\sin x}.\dfrac{1}{\sin
x}\]-----(4)
And as we know that, \[\dfrac{\cos x}{\sin x}=\cot x\] and \[\dfrac{1}{\sin x}=\csc x\], using these in
the equation (4), we get,
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{\sin x} \right)=-\cot x\csc x\]
Therefore, we get the differentiation of $f(x)=\csc x$ as:
\[f'(x)=-\cot x\csc x\].
Note: The pre-requisite of using the quotient rule is that the given expression should be in the form of fraction (with a numerator and numerator). And that is why wrote \[f(x)=\csc x=\dfrac{1}{\sin x}\]. The formula should be written correctly and the values should be substituted in the formula in different steps to minimize mistakes.
Complete step-by-step solution:
According to the given question, we are given an expression which we have to differentiate using the
quotient rule.
Quotient rule is a rule for differentiation of function which involves ratio of two functions which are
differentiable. The formula of the quotient rule is as follows.
\[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}(u)-u\dfrac{d}{dx}(v)}{{{v}^{2}}}\]
We will carry out the differentiation based on the above formula of quotient rule.
The given expression we have is,
$f(x)=\csc x$-----(1)
We can simplify it further and write the equation (1) as,
\[f(x)=\dfrac{1}{\sin x}\]-----(2)
Using the formula of quotient rule, we will substitute the values in it. And we have,
\[f'(x)=\dfrac{d}{dx}\left( \dfrac{1}{\sin x} \right)=\dfrac{\sin x\dfrac{d}{dx}(1)-1\dfrac{d}{dx}(\sin
x)}{{{\sin }^{2}}x}\]-----(3)
We proceed on to solve the above expression to get the differentiation value, we have,
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{\sin x} \right)=\dfrac{\sin x(0)-1(\cos x)}{{{\sin }^{2}}x}\]
And as we know that, differentiation of any constant gives us a value 0, that is, \[\dfrac{d}{dx}(1)=0\]
and differentiation of sine function is cosine function, that is, \[\dfrac{d}{dx}(\sin x)=\cos x\].
So, we get,
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{\sin x} \right)=\dfrac{-(\cos x)}{{{\sin }^{2}}x}\]
We can further proceed as,
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{\sin x} \right)=\dfrac{-\cos x}{\sin x}.\dfrac{1}{\sin
x}\]-----(4)
And as we know that, \[\dfrac{\cos x}{\sin x}=\cot x\] and \[\dfrac{1}{\sin x}=\csc x\], using these in
the equation (4), we get,
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{\sin x} \right)=-\cot x\csc x\]
Therefore, we get the differentiation of $f(x)=\csc x$ as:
\[f'(x)=-\cot x\csc x\].
Note: The pre-requisite of using the quotient rule is that the given expression should be in the form of fraction (with a numerator and numerator). And that is why wrote \[f(x)=\csc x=\dfrac{1}{\sin x}\]. The formula should be written correctly and the values should be substituted in the formula in different steps to minimize mistakes.
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