Learn definition of derivative as a function by first principle and find an expression for f?(x) by using definition of derivative. Practice differentiability of a function at a point & relationship between differentiability and continuity.
\(f'{(a)}\) \(=\lim\limits_{h\to 0}\;\left(\dfrac{f(a+h)-f(a)}{h}\right)\)
Now let \('a'\) be a variable number \(x\), so that as \(x\) changes this expression changes with \(x\) and hence becomes a function of \(x\).
\(f'{(x)}\) \(=\lim\limits_{h\to 0}\;\left(\dfrac{f(x+h)-f(x)}{h}\right)\)
We say that this is derivative of f( i.e., f'), provided the limit exists at that \(x\).
(There may be some values of \(x\) for which \(f'\) may not be defined or limit does not exist).
A \(\dfrac{1}{x+7}\)
B \(\dfrac{-2}{(2x+3)^2}\)
C \(\dfrac{5x}{x+3}\)
D \(x^2\)
We normally take \(y\) as the dependent variable and \(x\) as the independent variable.
\(y=f(x)\)
We use the notation:
derivatives of \(y\) with respect to \(x=f'(x)\)
Sometimes we use the notations,
\(f'(x)=y'=\underbrace{\dfrac{dy}{dx}=\dfrac{df}{dx}=\dfrac{df(x)}{dx}=Df(x)=D_xf(x)}_\text{All expression have the same meaning}\)
Note that \(f'{(a)}\) means first take the derivative of the function with respect to \(x\)and then put \(x=a\) in the derivative expression. It is not the derivative of \(f(a)\), which will always be 0 because \(f(a)\) is a constant.
A \(\dfrac{4}{(x+1)^3}\)
B \(\dfrac{2}{(x+1)^2}\)
C \(\dfrac{2x}{x^5+7}\)
D \(4x^5\)
i.e. if a function is continuous at \(x=a\), then it may or may not be differentiable at \(x=a\).
\(f(x)=|x|\) is non differentiable at \(x=0\), in general \(|g(x)|\) will be non differentiable at all the roots of \(g(x)=0\).
A \(f(x)=|x-7|\)
B \(f(x)=|x|\)
C \(f(x)=|x-9|\)
D \(f(x)=|x-3|\)
A function \('f'\) is said to be differentiable at \(x=a\) if \(f'{(a)}\) exists i.e the limit of definition of derivative exists.
i.e.
\(=\lim\limits_{h\to 0}\;\left(\dfrac{f(a+h)-f(a)}{h}\right)\) exists.