Derivatives

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

1. \(\frac{d}{dx} (sin~ x)= cos\ x\)
2. \(\frac{d}{dx} (cos~ x)= – sin\ x\)
3. \(\frac{d}{dx} (tan ~x)= sec^{2} x\)
4. \(\frac{d}{dx} (cot~ x = -cosec^{2} x\)
5. \(\frac{d}{dx} (sec~ x) = sec\ x\ tan\ x\)
6. \(\frac{d}{dx} (cosec ~x)= -cosec\ x\ cot\ x\)
7. \(\frac{d}{dx} (sinh~ x)= cosh\ x\)
8. \(\frac{d}{dx} (cosh~ x) = sinh\ x\)
9. \(\frac{d}{dx} (tanh ~x)= sech^{2} x\)
10. \(\frac{d}{dx} (coth~ x)=-cosech^{2} x\)
11. \(\frac{d}{dx} (sech~ x)= -sech\ x\ tanh\ x\)
12. \(\frac{d}{dx} (cosech~ x ) = -cosech\ x\ coth\ x\)

2. Formula for Inverse Trigonometric Functions

1. \(\frac{d}{dx}(sin^{-1}~ x)\) = \(\frac{1}{\sqrt{1 – x^2}}\)
2. \(\frac{d}{dx}(cos^{-1}~ x)\) = \(-\frac{1}{\sqrt{1 – x^2}}\)
3. \(\frac{d}{dx}(tan^{-1}~ x)\) = \(\frac{1}{1 + x^2}\)
4. \(\frac{d}{dx}(cot^{-1}~ x)\) = \(-\frac{1}{1 + x^2}\)
5. \(\frac{d}{dx}(sec^{-1} ~x) \)= \(\frac{1}{|x|\sqrt{x^2 – 1}}\)
6. \(\frac{d}{dx}(cosec^{-1}~x) \)= \(-\frac{1}{|x|\sqrt{x^2 – 1}}\)

3. Other Functions

1. \(\frac{d}{dx}(a^{x}) = a^{x} ln a\)
2. \(\frac{d}{dx}(e^{x}) = e^{x}\)
3. \(\frac{d}{dx}(log_a~ x)\) = \(\frac{1}{(ln~ a)x}\)
4. \(\frac{d}{dx}(ln~ x) = 1/x\)
5. Chain Rule: \(\frac{dy}{dx}\) = \(\frac{dy}{du} × \frac{du}{dx}\) = \(\frac{dy}{dv} × \frac{dv}{du} × \frac{du}{dx}\)