Question

Prove that the derivative of an odd function is always an even function.

Answer 1

Hint: Think of the basic definition of an odd function, i.e., f(-x)=-f(x) and also use the knowledge of differentiation to reach a result which can help you to conclude the fact asked in the question.

Complete step-by-step answer:
Let us start the solution to the above question by considering a general odd function $f(x)$ . As we have assumed that $f(x)$ is an odd function, so according to the definition of the odd function, we can write it mathematically as:
$f(-x)=-f(x)$
Now we know that the derivative of a function g(t) with respect to x is $g'(t)\dfrac{dt}{dx}$ . So, if we differentiate both sides of the above equation with respect to x, we get
$\dfrac{d\left( f(-x) \right)}{dx}=\dfrac{d\left( -f(x) \right)}{dx}$
$\Rightarrow f'(-x)\dfrac{d(-x)}{dx}=-f'(x)\dfrac{dx}{dx}$
Now we know that the value of \[\dfrac{d(-x)}{dx}=-1\text{ and }\dfrac{dx}{dx}=1\text{ }\] . So, if we put the value in our equation, we get
$-f'(-x)=-f'(x)$
Now, if we multiply both the sides of the equation by -1, we get
$f'(-x)=f'(x)$
Now looking at the above result, we can say that the function f’(x) is an even function and we know that f’(x) represents the derivative of f(x). So, we have proved that the derivative of an odd function is always an even function.

Note: While solving the problem related to function, we need to be very careful about the signs of the terms we are dealing with, as a small mistake of the sign would lead to a wrong answer. Also, you need to learn all the properties of odd and even functions, along with one-one, onto and other types of functions.