Question

How do you simplify \[\dfrac{1}{{\cos x}}?\]

Answer 1

Hint:We can try solving it by using the Quotient Rule. Quotient Rule can be applied here, because \[1\] over the cosine of \[x\] is a quotient. So, we will use it to find the derivatives. If two functions like \[f(x)\] and \[g(x)\] are there which are going to be differentiated then it requires the quotient rule.

Formula used:
\[{\left( {\dfrac{{f(x)}}{{g(x)}}} \right)^1} = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{\left( {g(x)} \right)}^2}}}\]

Complete step by step answer:
We will use the quotient rule that says:
\[{\left( {\dfrac{{f(x)}}{{g(x)}}} \right)^1} = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{\left( {g(x)} \right)}^2}}}\]
Now, to find the derivative of a quotient we will first find \[f(x)\] which is the function in the numerator and its derivative which is \[f'(x)\]. So, according to the formula we get that:
\[f(x) = 1\,\]
We know that \[1\] is constant and so its derivative is going to be \[0\]. So, \[f'(x) = 0\].Now, we need to find out the \[g(x)\]which is the function of the denominator here, and we have to also find its derivative which is \[g'(x)\]. So, here according to the formula we get that:
\[g(x) = \cos x\]
Now, we will find the derivative of \[g(x)\]which is \[g'(x)\]. So, we know that the derivative of \[\cos x\]is \[ - \sin x\]. Now, we will just put the values according to the formula, and then we will get:
\[{\left( {\dfrac{{f(x)}}{{g(x)}}} \right)^1} = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{\left( {g(x)} \right)}^2}}}\]
\[ \Rightarrow {\left( {\dfrac{{f(x)}}{{g(x)}}} \right)^1} = \dfrac{{\cos x \cdot 0 - 1 \cdot ( - \sin x)}}{{{{(\cos x)}^2}}}\]
\[ \Rightarrow {\left( {\dfrac{{f(x)}}{{g(x)}}} \right)^1} = \dfrac{{\sin x}}{{{{\cos }^2}x}}\]
Now, we will try to simplify this equation. According to the basic trigonometric functions, we know that:
\[\dfrac{1}{{\cos (x)}} = \sec (x)\]
\[\dfrac{{\sin (x)}}{{\cos (x)}} = \tan (x)\]
Now, we will just apply these basic formulas in our equation, and we will get:
\[ \Rightarrow \dfrac{{\sin x}}{{{{\cos }^2}x}} = \dfrac{{\sin x}}{{\cos x}} \times \dfrac{1}{{\cos x}}\]
\[ \Rightarrow \dfrac{{\sin x}}{{{{\cos }^2}x}} = \tan x\sec x\]
\[ \therefore {\left( {\dfrac{{f(x)}}{{g(x)}}} \right)^1} = \dfrac{{\sin x}}{{{{\cos }^2}x}} = \tan x\sec x\]

Therefore, we got the derivative of \[\dfrac{1}{{\cos x}}\] as \[\tan x\sec x\].

Note: We need to remember the basic formulas and the reciprocal formulas of trigonometry here. These are very important while solving problems like these. We use them when we calculate complex derivatives that need those formulas. We can also remember them because it would help us solve our question more easily and quickly.