Question

What is the derivative of $1 + {\tan ^2}x$?

Answer 1

Use chain rule of Differentiation to calculate the derivative of the given expression. Since $1$ is a constant, chain rule will practically be applied only on the trigonometric term of the expression.
Formula Used: Chain rule of Differentiation - $\dfrac{{d[f(g(x))]}}{{dx}} = f'(g(x)).g'(x)$

Complete step-by-step solution:
Derivative of a function is defined as the value/expression/function which is obtained when a function is differentiated. First Derivative refers to the value obtained when a function is differentiated once, Second Derivative refers to the value obtained when a function is differentiated twice in succession, Third Derivative is obtained when a function is differentiated three times in succession, and so on and so forth.
There are various methods used to differentiate different types of functions, where the choice of method depends on the complexity of the given function. One such type of function is a composite function. A function is said to be composite if it consists of a function within a function, and is expressed as $f(g(x))$, where $g(x)$ and $f(x)$ are both different and distinct functions and the composite function consists of $g(x)$ contained within $f(x)$.
To differentiate such composite functions, a very reliable method which is frequently used is the Chain Rule of Differentiation. This rule basically says that the composite function$f(g(x))$, which can be expressed as a function within a function, is differentiated such that the final derivative of the function is the product of derivative of $f(g(x))$ with respect to $g(x)$, multiplied by derivative of$g(x)$, that is, $\dfrac{{d[f(g(x))]}}{{dx}} = f'(g(x)).g'(x)$.
In the question given to us, the expression is $1 + {\tan ^2}x$, which can be expressed as a composite function where $f(x) = 1 + {(\tan x)^2}$ and $g(x) = x$.
Thus using the Chain rule of Differentiation on this expression we get,
$\dfrac{{d(1 + {{(\tan x)}^2})}}{{dx}} = 0 + 2\tan x\dfrac{{d(\tan x).}}{{dx}}\dfrac{{d(x)}}{{dx}}$
Since $1$ is a constant, its derivative will be $0$. Using the Chain rule, derivative of ${(\tan x)^2}$ will be exponent of $\tan x$, which is $2$, multiplied by $\tan x$, multiplied by the first derivative of $\tan x$, times the derivative of $g(x)$, which is $x$. We know that the derivative of $x$ is $1$. Hence, simplifying the equation we get,
$2\tan x\dfrac{{d(\tan x).}}{{dx}}1$
Now, to simplify this equation further, we need to find out the derivative of $\tan x$, and substitute its value in the equation.
\dfrac{{d(\tan x)}}{{dx}}$$$$ = {\sec ^2}x
Hence, substituting the value of $\dfrac{{d(\tan x)}}{{dx}}$ in the equation, we get,
\dfrac{{d(1 + {{(\tan x)}^2})}}{{dx}}$$$$ = 2\tan x({\sec ^2}x)
Hence, the derivative of $1 + {\tan ^2}x$ is $2\tan x{\sec ^2}x$.

Note: It is important to recognize whether the given function is composite or not before applying the Chain Rule. If wrongly applied, it may lead to wrong answers.