Question

find yn in the following case
(ii) y = 1/(ax + b )^m​

Answer 1

|| GIVEN ||

$\bf \: y = {(ax + b)}^{m}$

|| TO FIND ||

${n}^{th \:} derivative \: {y}^{n}$

|| SOLUTION ||

$y \: = {(ax + b)}^{m}$

Differentiate with respect to x ,

$\dfrac{dy}{dx} = \dfrac{d}{dx} {(ax + b)}^{m}$

$\implies {(ax + b)}^{m} = {m(ax + b)}^{m - 1}$

By substituting the above in dy/dx , we get

$= m {(ax + b)}^{m - 1} \times \dfrac{d}{dx} (ax + b)$

Using chain rule , So we had multiplied again with ax+b.

Now , We know that , always the variable takes the value 1 and the constant becomes 0 , Let's do substitution here.

$m {(ax + b)}^{m - 1} \times ((a \times 1) + 0)$

$\bf \: \dfrac{dy}{dx} = ma {(ax + b)}^{m - 1}$

We have found the first derivative , Now let's find the second derivative.

$\dfrac{ {d}^{2}y }{d {x}^{2} } = \dfrac{d}{dx} (ma( {ax + b)}^{m - 1})$

$= ma \dfrac{d}{dx} {(ax + b)}^{m - 1}$

$= ma \dfrac{d}{dx} {(ax + b)}^{m - 1}$

$= ma(m - 1) {(ax + b)}^{m - 2} \times \dfrac{d}{dx} (ax + b)$

Like the first derivative We must calculate for the second derivative.

$= ma(m - 1) {(ax + b)}^{m - 2} \\ \:\times (a \times 1 + 0)$

$m {a}^{2} (m - 1) {(ax + b)}^{m - 2}$

So, The nth derivative will be ,

$\dfrac{ {d}^{n}y }{d {x}^{n} } = {a}^{n}m(m - 1)(m - 2) ...(m - (n + 1)) \: {(ax + b)}^{m - n}$

______________________ .

Answer 2

Solution:-

 1st derivation

$\rm y_1=(ax+b)^{m-1} .m\dfrac{d}{dx} (ax+b)\\ \\\rm y_1 =a.m(ax+b)^{m-1}\\ \\\rm y_2 =a^{2}.m \: (m-1)(ax+b)^{m-2}\\ \\\rm y_3=a^{3}.m(m-1)(m-2)(ax+b)^{m-3} \\ \\\rm y_4=a^{4}.m(m-1)(m-2)(m-3)(ax+b)^{m-4}\\ \\\rm y_n=a^{n}.m{m-1}{m-2}{m-n+1}.{ax+b}^{m-n}$

If m = n

$y_n=a^{n}.m{m-1}{m-2}(ax+b)^{m-m}=1$

$\rm = a^{n}.m(m-1)(m-2)(m-3) _________ 3.2.1$

Answer,

       $\rm{y_n=a^{n}.m1}$