Derivatives
By Suresh Chand
The derivatives of a function f at a point a is denoted by f‘(a) and is defined as
\(f^{‘}(a) = \lim_{h \to 0} \frac{f(x) – f(a)}{x – a}\)
if limit exists
Suppose h = |x – a|. Then (1) can be written as
\(f^{‘}(a) = \lim_{h \to 0} \frac{f(a + h) – f(a)}{h}\)
Example: IF f(x) = x3 – x, then find a formula for f‘(x).
Solution:
Let f(x) = x3 – x
Then,
\(f^{‘}(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h}\)
\(= \lim_{h \to 0} \left ( \frac{(x + 3)^3 – (x + h) – x^3 + x}{h} \right )\)
\(= \lim_{h \to 0} \left ( \frac{x^3 + h^3 + 3x^2h + 3xh^2 – x – h – x^3 + x}{h} \right )\)
\(= \lim_{h \to 0} \left ( \frac{h(3x^2 + 3xh + h^2 – h)}{h} \right )\)
\(= \lim_{h \to 0} \left ( 3x^2 + 3xh + h^2 – h \right )\)
= 3x2 – 1
Thus, f‘(x) = 3x2 – 1
Show that f(x) = |x| is not differentiable at x = 0
Solution:
For x > 0, we have |x| = x. Therefore, for h > 0, |x + h| = x + h
Here, for > 0
\(f^{‘}(x) = \lim_{h \to 0}\left ( \frac{f(x + h) – f(x)}{h} \right )\)
\(= \lim_{h \to 0}\left ( \frac{|x + h| – |x|}{h} \right )\)
\(= \lim_{h \to 0}\left ( \frac{x + h – x} {h}\right )\)
\(= \lim_{h \to 0}\left ( \frac{h} {h}\right )\)
=1
This means f is differentiable for x > 0
and for x < 0, we have |x| = – x. Then choose h is so small such that |x + h| < 0 and |x + h| = -(x + h).
Here, for x < 0
\(f^{‘}(x) = \lim_{h \to 0}\left ( \frac{f(x + h) – f(x)}{h} \right )\)
\(= \lim_{h \to 0}\left ( \frac{-(x + h) – (-x)} {h}\right )\)
\(= \lim_{h \to 0}\left ( \frac{-h} {h}\right )\)
= -1
This means f is differentiable for x < 0
But,
\(f^{‘}(x)|_{\enspace for \enspace x > 0} \neq f^{‘}(x)|_{\enspace for \enspace x < 0}\)
So, f is not differentiable at x = 0
Geometrically, the curve f(x) doesn’t have a tangent line at origin i.e. at (0, 0).
Review of Derivative
1. Multiple Rule
If c be a constant and f is differentiable function then
\(\frac{d}{dx}(cf(x)) = c \frac{d}{dx}f(x)\)
2. Sum and difference rule
If f and g are both differentiable functions then
\(\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)\)
3. Product Rule
If f and g are both differentiable functions then
\(\frac{d}{dx}(f(x) \enspace g(x)) = f(x)\frac{d}{dx}g(x) + g(x)\frac{d}{dx}f(x)\)
Example 1: If f(x) = \(\frac{secx}{1 + tanx}\) then find f‘(x)
Solution
Let,
\(f(x) = \frac{secx}{1 + tanx}\)
Then
\(f^{‘}(x) = \frac{(1 + tanx) secx tanx – secx (sec^{2}x)}{(1 + tanx) ^ 2}\)
\(= \frac{secx (tanx + tan^2x – sec^2x)}{(1 + tanx) ^ 2}\)
\(= \frac{secx (tanx – 1)}{(1 + tanx) ^ 2}\)
Example 10: Find an equation of the tangent line to the curve y = 2x sinx at the point \((\frac{\pi}{2}, \pi)\).
Solution: Let,
y = 2x sinx
Then
y‘ = 2sinx + 2x cosx
At the point \((\frac{\pi}{2}, \pi)\),
\(y^{‘} = 2 sin(\frac{\pi}{2}) + 2\frac{\pi}{2}cos(\frac{\pi}{2})\)
= 2 + 0
= 2
This shows the curve has slope 2 at the point \((\frac{\pi}{2}, \pi)\). Since the point \((\frac{\pi}{2}, \pi)\), is the common point of the curve and the tangent line to y = 2x sinx at \((\frac{\pi}{2}, \pi)\). Therefore the slope of the tangent line is 2
Thus, the equation of tangent line to y = 2x sinx at the point \((\frac{\pi}{2}, \pi)\) is
\(y – \pi = 2(x – \frac{\pi}{2})\)
\(y – \pi = 2x – \pi\)
2x – y = 0
Differentiable Formulas
1. Formula for Trigonometric Functions
- \(\frac{d}{dx} (sin~ x)= cos\ x\)
- \(\frac{d}{dx} (cos~ x)= – sin\ x\)
- \(\frac{d}{dx} (tan ~x)= sec^{2} x\)
- \(\frac{d}{dx} (cot~ x = -cosec^{2} x\)
- \(\frac{d}{dx} (sec~ x) = sec\ x\ tan\ x\)
- \(\frac{d}{dx} (cosec ~x)= -cosec\ x\ cot\ x\)
- \(\frac{d}{dx} (sinh~ x)= cosh\ x\)
- \(\frac{d}{dx} (cosh~ x) = sinh\ x\)
- \(\frac{d}{dx} (tanh ~x)= sech^{2} x\)
- \(\frac{d}{dx} (coth~ x)=-cosech^{2} x\)
- \(\frac{d}{dx} (sech~ x)= -sech\ x\ tanh\ x\)
- \(\frac{d}{dx} (cosech~ x ) = -cosech\ x\ coth\ x\)
2. Formula for Inverse Trigonometric Functions
- \(\frac{d}{dx}(sin^{-1}~ x)\) = \(\frac{1}{\sqrt{1 – x^2}}\)
- \(\frac{d}{dx}(cos^{-1}~ x)\) = \(-\frac{1}{\sqrt{1 – x^2}}\)
- \(\frac{d}{dx}(tan^{-1}~ x)\) = \(\frac{1}{1 + x^2}\)
- \(\frac{d}{dx}(cot^{-1}~ x)\) = \(-\frac{1}{1 + x^2}\)
- \(\frac{d}{dx}(sec^{-1} ~x) \)= \(\frac{1}{|x|\sqrt{x^2 – 1}}\)
- \(\frac{d}{dx}(cosec^{-1}~x) \)= \(-\frac{1}{|x|\sqrt{x^2 – 1}}\)
3. Other Functions
- \(\frac{d}{dx}(a^{x}) = a^{x} ln a\)
- \(\frac{d}{dx}(e^{x}) = e^{x}\)
- \(\frac{d}{dx}(log_a~ x)\) = \(\frac{1}{(ln~ a)x}\)
- \(\frac{d}{dx}(ln~ x) = 1/x\)
- Chain Rule: \(\frac{dy}{dx}\) = \(\frac{dy}{du} × \frac{du}{dx}\) = \(\frac{dy}{dv} × \frac{dv}{du} × \frac{du}{dx}\)