Question

How do you find the derivative of $\dfrac{1}{{\cos x}}?$

Answer 1

Hint: As we know that the above given expression is a trigonometric expression as sine, cosine all are trigonometric ratios. Derivatives are defined as the varying rate of a function with respect to an independent variable. To find the derivative of the given function we simplify it using the quotient rule by using the trigonometric ratios. The quotient rule is: \[{\left[ {\dfrac{{f(x)}}{{g(x)}}} \right]^{'}} = \dfrac{{g(x)f{'}(x) - f(x)g{'}(x)}}{{{{\left[ {g(x)} \right]}^2}}}\]
We should first find the function in the numerator $f(x)$ and $g(x)$ then find their derivative $f{'}(x)$ and $g{'}(x)$.

Complete step by step answer:
Here the given function is $\dfrac{1}{{\cos x}}$. The numerator here is $1$, so $f(x) = 1$. Now we have to find the derivative of this function. Since $1$ is a constant and we know that the derivative of a constant is $0$. So $f{'}(x) = 0$.
The denominator in the function is $\cos x$ i.e. $g(x) = \cos x$. Now we know that the derivative of $\cos x$is $( - \sin x)$. So $g{'}(x) = - \sin x$. Now we will put the values in the quotient formula: \[{\left[ {\dfrac{{f(x)}}{{g(x)}}} \right]^{'}} = \dfrac{{g(x)f{'}(x) - f(x)g{'}(x)}}{{{{\left[ {g(x)} \right]}^2}}}\]
$ \Rightarrow \dfrac{{\cos x \times 0 - 1 \times ( - \sin x)}}{{{{(\cos x)}^2}}} = \dfrac{{\sin x}}{{{{\cos }^2}x}}$.
We can also write it as $\dfrac{1}{{\cos x}} \times \dfrac{{\sin x}}{{\cos x}}$,
Also we know that $\dfrac{1}{{\cos x}}$ is written as $\sec x$ and $\dfrac{{\sin x}}{{\cos x}}$ is written as $\tan x$.
Therefore we get $\sec x\tan x$.

Hence the required answer is $\sec x\tan x$.

Additional information:
\[ \bullet \]Linear combination rule: The linearity law is very important to emphasize its nature with alternate notation. Symbolically it is specified as \[h{'}(x) = af{'}(x) + bg{'}(x)\]
\[ \bullet \]Quotient rule: The derivative of one function divided by other is found by quotient rule such as\[{\left[ {\dfrac{{f(x)}}{{g(x)}}} \right]^{'}} = \dfrac{{g(x)f{'}(x) - f(x)g{'}(x)}}{{{{\left[ {g(x)} \right]}^2}}}\].
\[ \bullet \]Product rule: When a derivative of a product of two function is to be found, then we use product rule that is \[\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}\].
\[ \bullet \]Chain rule: To find the derivative of composition function or function of a function, we use chain rule. That is \[fog{'}({x_0}) = [(f{'}og)({x_0})]g{'}({x_0})\].

Note: We know the differentiation of \[{x^n}\] is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\]. The obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying the product rule. We also know that differentiation of constant terms is zero.