Revision:OCR Core 3 - Extending differentiation and integration

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Revision:OCR Core 3 - Extending differentiation and integration

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 3 - Extending differentiation and integration

1 4. Extending differentiation and integration
2 Also See
3 Comments

4. Extending differentiation and integration

Differentiating $(ax + b)^{n}$

Without the knowledge of information in Core 3, a student might wish to differentiate the aforementioned function using the rather cumbersome method of binomial expansion. For example:

$\frac{d}{dx} (x + 1)^{3} = \frac{d}{dx} (x^{3} + 3x^{2} + 3x + 1) = 3x^{2} + 6x + 3 = 3(x + 1)^{2}$ .

This might suggest a link however, and this can be further investigated using a general case:

$(ax + b)^{n} = {n \choose 0} a^{n}x^{n} + {n \choose 1} a^{n - 1}bx^{n - 1} + {n \choose 2} a^{n - 2}b^{2}x^{n - 2} + ... + {n \choose \left( n - 1 \right) } ab^{n - 1}x + {n \choose n}b^{n}$

Hence:

$\frac{d}{dx} (ax + b)^{n} = na^{n} {n \choose 0} x^{n - 1} + (n - 1)ba^{n - 1} {n \choose 1} x^{n - 2} + ... + ab^{n - 1} {n \choose \left( n - 1 \right) }$

A factor of "a" can be taken out, giving:

$\frac{d}{dx} (ax + b)^{n} = a \left( na^{n - 1} {n \choose 0} x^{n - 1} + (n - 1)ba^{n - 2} {n \choose 1} x^{n - 2} + ... + b^{n - 1} {n \choose \left( n - 1 \right) } \right)$

$a \left( na^{n - 1} {n \choose 0} x^{n - 1} + (n - 1)ba^{n - 2} {n \choose 1} x^{n - 2} + ... + b^{n - 1} {n \choose \left( n - 1 \right) } \right) \\ = a \left( n {n \choose 0} a^{n - 1} x^{n - 1} + (n - 1) {n \choose 1} ba^{n - 2} x^{n - 2} + ... + {n \choose \left( n - 1 \right) } b^{n - 1} \right)$

A factor of "n" can be taken from all the terms:

$na \left({ \left( n - 1 \right) \choose 0} a^{n - 1} x^{n - 1} + { \left(n - 1 \right) \choose 1} ba^{n - 2}x^{n - 2} + ... + { \left( n - 1 \right) \choose \left( n - 1 \right) } b^{n - 1} \right)$

Hence:

$\frac{d}{dx} (ax + b)^{n} = na(ax + b)^{n - 1}$ .

This can be extended to functions other than powers:

$\frac{d}{dx} f(ax + b) = af^{'}(ax + b)$ .

Integrating $(ax + b)^{n}$

Integration can also be covered using a similar rule:

$\int (ax + b)^{n} \,dx \\ = \int \left( {n \choose 0} a^{n}x^{n} + {n \choose 1} a^{n - 1}bx^{n - 1} + {n \choose 2} a^{n - 2}b^{2}x^{n - 2} + ... + {n \choose \left( n - 1 \right) } ab^{n - 1}x + {n \choose n}b^{n} \right) \,dx$

$\int (ax + b)^{n} \,dx = \left[ \frac{1}{n + 1} a^{n} {n \choose 0} x^{n + 1} + \frac{1}{n} ba^{n - 1} {n \choose 1} x^{n} + ... + b^{n} {n \choose n} x \right] + C$

$\left[ \frac{1}{n + 1} a^{n} {n \choose 0} x^{n + 1} + \frac{1}{n} ba^{n - 1} {n \choose 1} x^{n} + ... + b^{n} {n \choose n} x \right] \\ = \frac{1}{a} \left[ \frac{1}{n + 1} a^{n+1} {n \choose 0} x^{n + 1} + \frac{1}{n} ba^{n} {n \choose 1} x^{n} + ... + b^{n} {n \choose n} ax \right] \\ = \frac{1}{a} \left[ \frac{n!}{( n + 1)!} a^{n+1} x^{n + 1} + \frac{n!}{n!} ba^{n}x^{n} + ... + \frac{n!}{n!} b^{n}ax \right] \\ = \frac{1}{n + 1} \times \frac{1}{a} \left[ \frac{(n + 1)!}{(n + 1)!} a^{n+1} x^{n + 1} + \frac{(n + 1)!}{n!} ba^{n}x^{n} + ... + \frac{(n + 1)!}{n!} b^{n}ax \right] = \frac{1}{n + 1} \times \frac{1}{a} \left[ ax + b \right]^{n + 1}$

Again, this can be extended to other functions:

$\int f^{'}(ax + b) \,dx = \frac{1}{a} \times f(ax + b) + C$ .

Example

1. Find the area under the curve $y = (2x + 3)^{4}$ between x = 0 , and x = 1 .

$\int _{0} ^{1} (2x + 3)^{4} \,dx = \left[ \frac{1}{2} \times \frac{1}{5} \times (2x + 3)^{5} \right] ^{1} _{0}$ .

Hence, the area is:

$\left[ \frac{1}{10} (2x + 3)^{5} \right] ^{1} _{0} = \frac{3125 - 243}{10} = \frac{ 2882 }{10} = 288.2$ .

Justifying the rule for differentiating

The justification of the rules for the powers of functions does not apply for other functions of functions. We can consider the effects of the transformation from y = f(x) to y = f(ax + b) .

There is a stretch parallel to the x-axis, scale factor $\frac{1}{a}$ , and then a translation in the negative x-direction of "b" units.

Quite obviously the gradient of the curve is not affected by translations, however stretches are different. This stretch causes the gradient to increase by a factor of "a". (In cases with negative "a", there might need to be an extra transformation, but this is easy to deal with, and produces the same result).