How do you differentiate $f(x)=\csc x$ using the quotient rule?
Answer
Verified615.9k+ views
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.
Recently Updated Pages
What is BLO What is the full form of BLO class 8 social science CBSE
Explain the Treaty of Vienna of 1815 class 10 social science CBSE
10 examples of friction in our daily life
Draw a diagram of nephron and explain its structur class 11 biology CBSE
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE
A Paragraph on Pollution in about 100-150 Words
Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE
Which are the Top 10 Largest Countries of the World?
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE
Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE
What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?