powered by Tutorvista.com

Sales Toll Free No: 1-855-666-7440

The derivative of a function is found by differentiating the function with respect to a given variable. A function may have different combinations of expressions involving the variable. To differentiate such functions, there are rules to help us. Such rules are called as **derivative rules**. There are a number of derivative rules which have been framed using the first principles.These rules are like formulas and are applied in the differentiation of different types of functions. Let us study some important derivative rules in the following sections.

The product rule states that the derivative of a product of two functions is the sum of the derivative of the first function times the second function and the derivative of the second function times the first function. That is if u and v are two functions of the same variable, then,

(u$\times$v)' = u'$\times$v + v'$\times$u The quotient rule states that the derivative of a quotient of two functions is the difference of the derivative of the numerator function times the denominator function and the derivative of the denominator function times the numerator function, the whole thing divided by the square of the denominator function. That is, if u and v are two functions of the same variable, then,

($\frac{u}{v}$)' = $\frac{(u' \times v - v' \times u)}{(v^{2})}$, v $\neq$ 0 This is a very important derivative rule that helps us in finding the derivative of a composite function. If a function is another function of the variable then the derivative of the whole function with respect to the variable is the product of the derivative of the main function with respect to the inner function and the derivative of the inner function with respect to the variable. That is, if y is a function of u and u is a function of x, then,

$(\frac{dy}{dx})$ = $(\frac{dy}{du})\times(\frac{du}{dx})$Some difficult functions, even though straight may be difficult to differentiate directly. In such cases it is easier to make the given function into a composite function by suitable substitution and then differentiate them using the chain rule. This method of differentiation is called the differentiation by the method of substitution.

$(\frac{dy}{dx})$ = $(\frac{dy}{du})\times(\frac{du}{dx})$Some difficult functions, even though straight may be difficult to differentiate directly. In such cases it is easier to make the given function into a composite function by suitable substitution and then differentiate them using the chain rule. This method of differentiation is called the differentiation by the method of substitution.

If, u(x) = sin (x) , u(x + h) = sin (x + h). Using the sum formula for sine,

u(x + h) = sin (x)$\times$cos(h) + cos (x)$\times$sin (h)

Therefore, u(x + h) – u(x) = sin (x)$\times$cos(h) + cos (x)$\times$sin (h) – sin (x) = sin (x)[cos (h) – 1] + cos (x)$\times$sin (h)

Dividing by h both sides,

$\frac{[u(x + h) – u(x)]}{(h)}$ = $[\frac{\sin (x)\times{\cos (h) – 1}}{(h)}] + [\sin (x)\times{\frac{\sin (h)}{(h)}}]$

Now, as per the definition of a derivative of a function,

u '(x) = $\lim_{ h \to 0}$ $\frac{[u(x + h) – u(x)]}{(h)}$

= $lim_{h \to 0}$ $[\frac{sin(x)\times{cos(h) – 1}}{(h)}] + [\sin (x)\times{\frac{sin (h)}{(h)}}]$

= sin (x)* 0 + cos (x)(1) = cos (x)

Therefore, the derivative of sin (x) is, cos (x)

Now let, v(x) = cos (x)

This can be written as v(x) = sin [$\frac{\pi}{2}$ – x] = sin ($\alpha$), where $\alpha$ = [$\frac{\pi}{2}$ – x]

Now using chain rule, v'(x) = $(\frac{dv}{d\alpha})\times(\frac{d\alpha}{dx})$ = cos ($\alpha$)$\times$(-1) = -cos [$\frac{\pi}{2}$ – x] = -sin (x)

Therefore, the derivative of cos (x) is, – sin (x)

Let us now say that f(x) = tan(x) =$\frac{\sin (x)}{\cos (x)}$ = $\frac{u(x)}{v(x)}$

As per quotient rule, f '(x) = $\frac{[v(x) \times u'(x) – v'(x) \times u(x)]}{[v(x)]^{2}}$

= $\frac{[\cos (x)\times\cos' (x) – (- \sin x)\times\sin (x)]}{[cos (x)]^{2}}$

= $\frac{[\cos^{2} (x) + \sin^{2} (x)]}{[cos (x)]^{2}}$

= $\frac{ [1]}{[cos (x)]^{2}}$

= $\sec^{2} (x)$

Therefore, the derivative of tan (x) is, $\sec^{2} (x)$

Let us look into the derivative of f(x) = csc (x). This function can be written in the form $\frac{u(x)}{v(x)}$, where,

u(x) = 1 and v(x) = sin (x)

As per quotient rule, f '(x) =$\frac{[v(x) \times u'(x) – v'(x) \times u(x)]}{[v(x)]^{2}}$

= $\frac{[\sin (x)\times 0 – 1\times (\cos (x)]}{[\sin (x)]^{2}}$

= $\frac{[-\cos (x)]}{[\sin^{2} (x)]}$

= $\frac{[-\cos (x)]}{[\sin (x)\times\sin (x)]}$

= -csc (x)$\times$cot (x)

Therefore, the derivative of csc (x) is, -csc (x)$\times$cot (x)

The derivative of f(x) = sec (x) can be figured out as follows.

f(x) = sec (x) = csc [$\frac{\pi}{2}$ – x] = csc ($\alpha$), where $\alpha$ = [$\frac{\pi}{2}$ – x]

Now using chain rule,

f'(x) = ($\frac{df}{d\alpha}$)$\times$($\frac{d\alpha}{dx}$) = -csc ($\alpha$)$\times$cot ($\alpha$)$\times$(-1) = csc [$\frac{\pi}{2}$ – x]$\times$cot [$\frac{\pi}{2}$ – x] = sec (x)$\times$tan (x)

Therefore, the derivative of sec (x) is, sec (x)$\times$tan (x)

The derivative of f(x) = cot (x) can be figured out as follows.

f(x) = cot (x) = tan [$\frac{\pi}{2}$ – x] = tan ($\alpha$), where $\alpha$= [$\frac{\pi}{2}$– x]

Now using chain rule,

f'(x) = ($\frac{df}{d\alpha}$)$\times$($\frac{d\alpha}{dx}$) = $sec^{2} (\alpha)\times$(-1) = - sec

Therefore, the derivative of cot (x) is, $ csc^{2} (x)$

For the purpose of establishing the basic trig derivative rules, we assumed the functions as direct ratios of sine, cosine tangent, etc. But even if the functions are of composite nature, this rule is still applicable and the chain rule must be used in addition.

f(x + h) = log (x + h), where h is the infinitesimal increment in the value of the variable.

f(x + h) – f(x) = log (x + h) – log (x). Using the quotient rule of logarithm,

= log $[\frac{(x +h)}{(x)}]$ = log [1 +($\frac{h}{x}$)]

Since the ratio ($\frac{h}{x}$) is a very small fraction (as h is only a infinitesimal increment of x) , we can expand

log [1 +($\frac{h}{x}$)] = [$\frac{h}{x}$] – [$\frac{(\frac{h}{x})^{2}}{2}$] + [$\frac{(\frac{h}{x})^{3}}{3}$] – [$\frac{(\frac{h}{x})^{4}}{4}$]+ ………. Now dividing both sides by h,

log $\frac{[1 +(\frac{h}{x})]}{[h]}$ = [$\frac{1}{x}$] – [$\frac{(h)}{2(x)^{2}}$] + [$\frac{(h^{2})}{3(x)^{3}}$] – [$\frac{(h^{3})}{4(x)^{4}}$] ……….

= [($\frac{1}{x}$) – (or +) the terms containing powers of h].

The left side is nothing but the difference quotient of f(x) = log (x) and therefore, when h tends to 0, all the terms on the right become 0 excepting the first term ($\frac{1}{x}$).There fore, the derivative of ln (x) is, $\frac{1}{x}$.

Now let us extend the concept to find the derivative of a logarithmic function with any base.

Let f(x) =$\log_{b} (x)$

As per the change of the base of rule of logarithms,

[$\log_{b} (x)$]$\times$[$\log (b)$] = log (x)

or, f(x) $\times$ [log (b)] = log (x)

Now taking derivative on both sides with respect to x and since [log (b)] is a constant,

f '(x) $\times$ [log (b)] = [$\frac{1}{(x)}$]

or, f'(x) = $\frac{[1]}{[x\times\log (b)]}$

Therefore, the derivative of $\log_{b} (x)$ is, $\frac{[1]}{[x\times\log (b)]}$

Even if the logarithmic functions are of composite nature, this rule is still applicable and the chain rule must be used in addition.

Taking logarithm on both sides to a natural base,

log [f(x)] = log ($a^{x}$)

or, log [f(x)] = x$\times$log (a)

Now differentiating both sides with respect to x (this method is also called as implicit differentiation),

[$\frac{1}{f(x)}$]$\times$ f'(x) = log (a)

or, f'(x) = f(x) $\times$ log (a)

or, f'(x) = $a^{x}$ $\times$log (a)

Thus, the derivative of $a^{x}$ = $a^{x}\times$ log (a).

Even if the exponent is any other expression in x, the derivative can be found by the chain rule.

As a special case, if a = e, the exponential constant, then, plugging in a = e,

the derivative of $e^{x}$ = $e^{x}\times$ log (e) = $e^{x}$ $\times$1 = $e^{x}$.

Thus, we get an important result that the derivative of $e^{x}$ = $e^{x}$ itself.

That is, if a function is denoted as z = f(x, y), the derivative of the function with respect to variable x is called the partial derivative of the function with respect to x and denoted as ($\frac{\partial z}{\partial x}$) or $f_{x}$.

Same is the case if the derivative is with respect to variable y and denoted as ($\frac{\partial z}{\partial y}$) or $f_{y}$.

When you take the partial derivative with respect to x, the variable y is considered as constant and vice versa.

**For Example,** if z = f(x, y) = $f_{x}$ = x^{2} + 2xy, then ($\frac{\partial z}{\partial x}$) = $f_{x}$ = 2x + 2y, whereas, ($\frac{\partial z}{\partial y}$) = $f_{y}$ = 2x

Same is the case if the derivative is with respect to variable y and denoted as ($\frac{\partial z}{\partial y}$) or $f_{y}$.

When you take the partial derivative with respect to x, the variable y is considered as constant and vice versa.

Like in case of total derivatives, higher order partial derivatives are evaluated by differentiating the first derivative again. That is,

($\frac{\partial^{2} z}{\partial x^{2}}$) = [$\frac{\partial z(\frac{\partial z}{\partial x})}{\partial x}$] and ($\frac{\partial^{2} z}{\partial y^{2}}$) = [$\frac{\partial z(\frac{\partial z}{\partial y})}{\partial y}$]

Also the partial derivative of the first derivative with respect to the second variable is also a higher order partial derivative. That is, ($\frac{\partial^{2} z}{\partial x^{2}}$) = [$\frac{\partial z(\frac{\partial z}{\partial x})}{\partial x}$] and ($\frac{\partial^{2} z}{\partial y^{2}}$) = [$\frac{\partial z(\frac{\partial z}{\partial y})}{\partial y}$]

[$\frac{\partial z(\frac{\partial z}{\partial y})}{\partial y}$] = ($\frac{\partial^{2} z}{\partial x \partial y}$) and [$\frac{\partial z(\frac{\partial z}{\partial y})}{\partial x}$]= ($\frac{\partial^{2} z}{\partial y \partial x}$) are also second order partial derivatives. In case of defined and continuous functions, ($\frac{\partial^{2} z}{\partial x \partial y}$) = ($\frac{\partial^{2} z}{\partial y \partial x}$)