Definition of Derivative
If \( f(x) \) is a function of \( x \), a is the point in its domain, the derivative of \( f(x) \) at \( a \) is given by the limit:
\[
f'(a) = \lim_{{h \to 0}} \frac{f(a+h) - f(a)}{h}
\]
provided this limit exists.
This definition of derivative is also called the first principle of derivative
Derivative of Real Valued Function
If \( y \) is a function of \( x \), the derivative of \( y \) with respect to \( x \) is given by the limit:
\[
\frac{dy}{dx} = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}
\]
(a) It is present if this limit exists.
(b)This definition of derivative is also called the first principle of derivative
(c)The domain of definition of $\frac{dy}{dx}$ is wherever the above limit exists.
(d)There are different notations for derivative of a function
$\frac{dy}{dx}$, $\frac{df(x)}{dx}$
Example
$y=x^3$
\[
\frac{dy}{dx} = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}= \lim_{{h \to 0}} \frac{(x+h)^3 - x^3}{h}
\]
$= \lim_{{h \to 0}} \frac{x^3+3x^2h+3xh^2+h^3 - x^3}{h}= \lim_{{h \to 0}} {3x^2+3xh+h^2}= 3x^2$
Geometric Interpretation
From a geometric viewpoint, the derivative at a point on the curve of the function represents the slope of the tangent to the curve at that point.
Algebra of derivatives
Multiplication by Constant
$\frac {d}{dx} [cf(x)] = c \frac {d}{dx} f(x) $
Example
$\frac {d}{dx} [2 sinx ] = 2 \frac {d}{dx} sin x =2 cos (x)$
Addition and Subtraction
$\frac {d}{dx} [f(x)+g(x)]=\frac {d}{dx} f(x) + \frac {d}{dx} g(x)$
$\frac {d}{dx} [f(x)-g(x)]=\frac {d}{dx} f(x) - \frac {d}{dx} g(x)$
Example
$\frac {d}{dx} [sinx -cos x ] = \frac {d}{dx} sin x - \frac {d}{dx} cos x =cos (x) + sin(x)$
Multiplication
$\frac {d}{dx} [f(x)g(x)]=g(x) \frac {d}{dx} f(x) + f(x) \frac {d}{dx} g(x)$
Example
$\frac {d}{dx} [x^2 sinx ] = x^2 \frac {d}{dx} sin x + sin x \frac {d}{dx} x^2 =x^2 cos (x) + 2x sin(x)$
Division
$\frac {d}{dx} [f(x)/g(x)]=\frac {g(x) \frac {d}{dx} f(x) - f(x) \frac {d}{dx} g(x)}{[g(x)]^2} $
Example
$\frac {d}{dx} [sin(x) /x^2]=\frac {x^2 \frac {d}{dx} sin(x) - sin(x) \frac {d}{dx} x^2}{x^4} $
$=\frac {x^2 cos (x) - 2x sin(x)}{x^4}$
$=\frac {x cos (x) - 2 sin(x)}{x^3}$
Chain Rule
if y = f(u) and u =g(x) ,then
$\frac {dy}{dx} = \frac {dy}{du} \frac {du}{dx}$
Example
$\frac {d}{dx} [sin (x^3)] = \frac {d}{du} sin (u) \frac {d}{dx} (x^3)= 3x^2 cos (x^3)$
Standard Differentiation formulas and Proofs
(I)$\frac {d}{dx} (c) = 0$ ( Where c is a constant)
Proof
\[
\frac{dy}{dx} = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}= \lim_{{h \to 0}} \frac{c - c}{h}= \lim_{{h \to 0}} 0=0
\]
(II)$\frac {d}{dx} (cx) = c$ ( Where c is a constant)
Proof
\[
\frac{dy}{dx} = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}= \lim_{{h \to 0}} \frac{c(x+h) - cx}{h}= \lim_{{h \to 0}} \frac{ch}{h} =\lim_{{h \to 0}} c = 0
\]
(III)$\frac {d}{dx} (x^n) = nx^{n-1}$
Proof
\[
\frac{dy}{dx} = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}= \lim_{{h \to 0}} \frac{(x+h)^n -x^n}{h}
\]
Now By Binomial Theorem
$(x+h)^n =x^n + nx^{n-1) h + ...+ h^n$
Hence
$(x+h)^n -x^n=nx^{n-1) h + ...+ h^n$
Therefore
$=\lim_{{h \to 0}} \frac{nx^{n-1) h + ...+ h^n}{h}= \lim_{{h \to 0}} (nx^{n-1) + ...+ h^{n-1}) = nx^{n-1}$
Derivatives of Trigonometric Functions
Important formula to find the derivative
$\frac {d}{dx} f(x) =\displaystyle \lim_{h \to 0} \frac {f (x+h) - f(x)}{h}$
$\displaystyle \lim_{x \to 0} \frac {sinx}{x}=1$
Derivative of Sine Function
$\frac {d}{dx} (sin x) = cos x $
Proof of sin derivative
$\frac {d}{dx} (sin x) = \displaystyle \lim_{h \to 0} \frac {sin (x+h) - sin x}{h}= \displaystyle \lim_{h \to 0} \frac {2cos (\frac {2x+h}{2}) sin \frac {h}{2}}{h}$
$=\displaystyle \lim_{h \to 0} cos (x + \frac {h}{2}) \displaystyle \lim_{h \to 0} \frac {sin \frac {h}{2}}{\frac {h}{2}} = cos x$
as $\displaystyle \lim_{x \to 0} \frac {sinx}{x}=1$
Derivative of Cos Function
$\frac {d}{dx} (cos x) = -sin x $
Proof of Cos derivative
$\frac {d}{dx} (cos x) = \displaystyle \lim_{h \to 0} \frac {cos (x+h) - cos x}{h}= \displaystyle \lim_{h \to 0} -\frac {2sin (\frac {2x+h}{2}) sin \frac {h}{2}}{h}$
$=-\displaystyle \lim_{h \to 0} sin (x + \frac {h}{2}) \displaystyle \lim_{h \to 0} \frac {sin \frac {h}{2}}{\frac {h}{2}} = -sin x$
as $\displaystyle \lim_{x \to 0} \frac {sinx}{x}=1$
Derivative of Tan Function
$\frac {d}{dx} (tan x) = sec^2x $
Proof of tan derivative
$\frac {d}{dx} (tan x) = \displaystyle \lim_{h \to 0} \frac {tan (x+h) - tan x}{h}= \displaystyle \lim_{h \to 0} \frac {1}{h}[\frac {sin(x+h)}{cos(x+h) }- \frac {sin(x)}{cos(x)}]$
$=\displaystyle \lim_{h \to 0} \frac {1}{h}\frac {sin(x+h)cos (x) - sin(x) cos (x+h)}{cos (x+h) cos x}$
$=\displaystyle \lim_{h \to 0} \frac {1}{h} \frac {sin (x+h -x)}{cos (x+h) cos x} $
$=\displaystyle \lim_{h \to 0} \frac {sin h}{h} \displaystyle \lim_{h \to 0} \frac {1}{cos (x+h) cos x}= sec^2 x$
as $\displaystyle \lim_{x \to 0} \frac {sinx}{x}=1$
Derivative of Cot Function
$\frac {d}{dx} (cot x) = -cosec^2x $
Derivative of Secant Function
$\frac {d}{dx} (sec x) = sec x . tan x $
Derivative of Cosecant Function
$\frac {d}{dx} (cosec x) = -cosec x . cot x$
Derivatives of Exponential Functions
$\frac {d}{dx} (e^x) = e^x$
$\frac {d}{dx} (ln x) = \frac {1}{x}$
$\frac {d}{dx} (log_{10} x) =\frac {1}{x ln 10}$
$\frac {d}{dx} (log_{a} x) =\frac {1}{x ln a}$
$\frac {d}{dx} (a^x) = a^x ln a$
Related Topics Also Read
Notes
NCERT Solutions & Assignments
Please enable JavaScript to view the comments powered by Disqus.
