Question

How do you use the quotient rule to find the derivative of $y = \tan x$ ?

Answer 1

In the given problem, we are required to differentiate $y = \tan x$ with respect to x. Since, $y = \tan x$ can be represented as a rational function, so we will have to apply a quotient rule of differentiation in the process of differentiating rational functions involving the trigonometric functions. Also derivatives of basic algebraic and trigonometric functions must be remembered thoroughly.

Complete step by step answer:
To find the derivative of $y = \tan x$ with respect to x using the quotient rule of differentiation, we have to first represent the expression as a rational function.
We know that tangent is the ratio of sine and cosine. So, $\tan x = \dfrac{{\sin x}}{{\cos x}}$
So, Derivative of $y = \dfrac{{\sin x}}{{\cos x}}$ with respect to $x$can be calculated as $\dfrac{d}{{dx}}\left( {\dfrac{{\sin x}}{{\cos x}}} \right)$ .
Now, $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{\sin x}}{{\cos x}}} \right)$ .
Now, using the quotient rule of differentiation, we know that $\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right) - f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right)}}{{{{\left[ {g(x)} \right]}^2}}}$
So, Applying quotient rule to $\dfrac{d}{{dx}}\left( {\dfrac{{\sin x}}{{\cos x}}} \right)$, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos x \times \dfrac{d}{{dx}}\left( {\sin x} \right) - \sin x \times \dfrac{d}{{dx}}\left( {\cos x} \right)}}{{{{\left[ {\cos x} \right]}^2}}}$
Substituting the derivative of $\sin x$ as $\cos x$ and derivative of $\cos x$ as $- \sin x$, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos x \times \cos x - \sin x \times \left( { - \sin x} \right)}}{{{{\left[ {\cos x} \right]}^2}}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\left[ {\cos x} \right]}^2}}}$
Now, we know that ${\cos ^2}x + {\sin ^2}x = 1$. So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\cos }^2}x}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = {\sec ^2}x$
So, the derivative of the function $y = \tan x$ is ${\sec ^2}x$.

Note: The given problem may also be solved using the first principle of differentiation. The derivatives of basic trigonometric and algebraic functions must be learned by heart in order to find derivatives of complex composite functions using product rule and chain rule of differentiation.