Question

$\LARGE \boxed{ \underline{\bf \purple{ QUESTION :-}}}$


Find the nth derivative of $\sf \sqrt{ax + b}$​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:y = \sqrt{ax + b}$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} \sqrt{ax + b}$

$\rm :\longmapsto\:y_1 = \dfrac{1}{2} { \bigg(ax + b \bigg)}^{ \dfrac{1}{2} - 1}\dfrac{d}{dx}(ax + b)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}\bigg \}}$

$\red{\rm :\longmapsto\:y_1 = {( - 1)}^{0} \times \dfrac{a}{2} { \bigg(ax + b \bigg)}^{ - \dfrac{1}{2}}}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y_1 =\dfrac{d}{dx} \dfrac{a}{2} { \bigg(ax + b \bigg)}^{ - \dfrac{1}{2}}$

$\rm :\longmapsto\:y_2 = - \dfrac{1}{2} \times \dfrac{a}{2} { \bigg(ax + b \bigg)}^{ - \dfrac{1}{2} - 1} \times a$

$\red{\rm :\longmapsto\:y_2 = {( - 1)}^{1} \times \dfrac{ {a}^{2} }{ {2}^{2} } { \bigg(ax + b \bigg)}^{ - \dfrac{3}{2}}}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y_2 = \dfrac{d}{dx} \bigg( - \dfrac{ {a}^{2} }{ {2}^{2} } { \bigg(ax + b \bigg)}^{ - \dfrac{3}{2}} \bigg)$

$\rm :\longmapsto\:y_3 = - \dfrac{ {a}^{2} }{ {2}^{2} } \times \dfrac{( - 3)}{2} { \bigg(ax + b \bigg)}^{ - \dfrac{3}{2} - 1} \times a$

$\red{\rm :\longmapsto\:y_3 = {( - 1)}^{2} \dfrac{ 3 \times 1 \times {a}^{3} }{ {2}^{3} } { \bigg(ax + b \bigg)}^{ - \dfrac{5}{2}}}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y_3 =\dfrac{d}{dx} \bigg({( - 1)}^{2} \dfrac{ 3 \times 1 \times {a}^{3} }{ {2}^{3} } { \bigg(ax + b \bigg)}^{ - \dfrac{5}{2}} \bigg)$

$\rm :\longmapsto\:y_4 = {( - 1)}^{2} \dfrac{1 \times 3 \times {a}^{3} }{ {2}^{3} } \times \dfrac{( - 5)}{2} { \bigg(ax + b \bigg)}^{ - \dfrac{5}{2} - 1} \times a$

$\red{\rm :\longmapsto\:y_4 = {( - 1)}^{3} \dfrac{1 \times 3 \times \times 5 \times {a}^{4} }{ {2}^{4} }{ \bigg(ax + b \bigg)}^{ - \dfrac{7}{2}}}$

So on make out the pattern, we get nth derivative is

$\red{\rm :\longmapsto\:y_n = {( - 1)}^{n - 1} \dfrac{{a}^{n} }{ {2}^{n} }(1 \times 3 \times - - \times (2n - 3)) { \bigg(ax + b \bigg)}^{ - \dfrac{(2n - 1)}{2}}}$

Answer 2

$\large\fbox\colorbox{pink}{❥Answer}$

$\sf \frac{dy}{dx} = \frac{d}{dx} \sqrt{ax + b}$

$\sf y_1 = \frac{1}{2} (ax + b)^{ \frac{1}{2} - 1 } \frac{d}{dx} (ax + b)$

$\sf y_1 = ( - 1)^{0} \times \frac{a}{2} (ax + b)^{ \frac{ - 1}{ \: \: 2} }$

$\sf \frac{d}{dx} y_1 = \frac{d}{dx} \frac{a}{2} (ax + b)^{ \frac{ - 1}{ \: \: 2} }$

$\sf y_2 = \frac{ - 1}{ \: \: 2} \times \frac{a}{2} (ax + b)^{ \frac{ - 1}{ \: \: 2} - 1} \times a$

$\sf y_2 = { (- 1)}^{1} \times \frac{ {a}^{2} }{ {2}^{2} } (ax + b)^{ \frac{3}{2} }$

$\sf \frac{d}{dx} y_2 = (\frac{ - a^{2} }{ \: \: {2}^{2} } (ax + b)^{ \frac{ - 3}{ \: \: 2} } )$

$\sf y_3 = \frac{ - {a}^{2} }{ \: \: {2}^{2} } \times \frac{ - 3}{ \: \: 2} (ax + b)^{ \frac{ - 3}{ \: \: 2} - 1} \times a$

$\sf y_3 = ( - 1)^{2} \frac{3 \times 1 \times {a}^{2} }{2} (ax + b)^{ \frac{ - 5}{ \: \: 2} }$

$\sf \frac{d}{dx} y_3 = \frac{d}{dx} ( { - 1)}^{2} \frac{3 \times 1 \times {a}^{2} }{ {2}^{3} } (ax + b)^{ \frac{ - 5}{ \: \: 2} } )$

$\sf y_4 = { (- 1)}^{2} \frac{1 \times 3 \times {a}^{2} }{ {2}^{3} } \times \frac{ - 5}{ \: \: 2} (ax + b)^{ \frac{ - 5}{ \: \: 2} - 1} \times a$

$\sf y_4 = {( - 1)}^{3} \frac{1 \times 3 \times 5 \times {a}^{4} }{ {2}^{4} } (ax + b)^{ \frac{ - 9}{ \: \: 2} }$

We get nth derivative :-

$\sf \red{yn = {( - 1)}^{n - 1} \frac{ {a}^{n} }{ {2}^{n} } (1 \times 3 \times - \times (2n - 3)(ax + b)^{ \frac{ - (2n - 1)}{2} } }$

$\tiny\tt{ Hope \: this \: helps \: don't \: report }$