Difference Quotient Examples Difference Quotient Examples The difference quotient is used in the derivative. Dividing function difference from the difference of point is called as the difference quotient it is otherwise known as Newton's quotient. X and Y are the two distinct points on the graph of function f. A line passing through the 2 points X (x, f(x)) and Y ((x + h), f(x + h)), the formula to find difference quotient is [f(x+h)−f(x)]h The skill to set up and simplify difference quotients is a necessary help for calculus students.This is from the difference quotient that the basic formulas for derivatives are developed. X and Y are points on the graph of f. A line passing all the way through the 2 points X (x, f(x)) and Y(x+h, f(x+h)) is called a secant line. The derivative of (4x - 2)/(x^2 + 1) is: Know More About Partial Differential Equation
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\begin{align}\frac{d}{dx}\left[\frac{(4x - 2)}{x^2 + 1}\right] &= \frac{(x^2 + 1)(4) - (4x - 2)(2x)} {(x^2 + 1)^2}\\ & = \frac{(4x^2 + 4) - (8x^2 - 4x)}{(x^2 + 1)^2} = \frac{-4x^2 + 4x + 4}{(x^2 + 1)^2}\end{align} In the example above, the choices g(x) = 4x - 2 h(x) = x^2 + 1 were made. Analogously, the derivative of sin(x)/x2 (when x Ă˘â€°  0) is: \frac{\cos(x) x^2 - \sin(x)2x}{x^4} Another example is: f(x) = \frac{2x^2}{x^3} whereas g(x) = 2x^2 and h(x) = x^3, and g'(x) = 4x and h'(x) = 3x^2. The derivative of f(x) is determined as follows: f'(x) = \frac {\left(4x \cdot x^3 \right) - \left(2x^2 \cdot 3x^2 \right)} {\left(x^3\right)^2} = \frac{4x^4 - 6x^4}{x^6} = \frac{-2x^4}{x^6} = -\frac{2}{x^2} This can be checked by using laws of exponents and the power rule: f(x) = \frac{2x^2}{x^3} = \frac{2}{x} = 2x^{-1} f'(x) = -2x^{-2} = -\frac{2}{x^2}
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Linear Approximation Formula Linear Approximation Formula Linear approximation helps us to estimate the difficult to calculate type curved graph. It uses the values on the line which we want; this point is very close to the graph. That is gives the almost desired values of the point. Linear approximation helps in many numerical techniques such as Newton Raphson method to approximate the values. Let’s move on to the formula to compute the linear approximation. Let 'x' be any value lies in the range of any function say f(x). Tangent line at the point (a0, b0) on the graph of function f(x) has the equation. Start with the linear approximation formula, L(x) = f(a) + (x-a)f'(a). We need to substitute for a, f(a) and f'(a). We are told that a = 1 (that is the point near which we are approximating the logarithm). Thus, f(a) = ln(1) = 0.
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And f'(x) = 1/x, so that f'(a) = 1/1 = 1. Substituting all of this in the formula gives L(x)= f(a) + (x - a)f'(a) = 0 + (x - 1)(1) = x – 1. We have negative percentage which represents that the approximation is greater than the actual value. Example 2: Compute the linear approximation of p(x) = x sin( Π x2 ) at x = 2. You can use p( 1.99 ). Solution: formula: L(x) = f( a ) + f’( a )( x – a ), Here, a = 2 so p( 2 ) and p’(2),
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Thank You
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