Question

find the second derivative sin(ax+b)​

Answer 1

Solution :

$\sf{Function\:of\:y = sin(ax + b)}$

$\textsf{By applying the chain rule of differentiation and substituting the values in it,}\\\textsf{we get :-}$

$\underline{\sf{Chain\:rule\: differentiation :-}} \\ \\ \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}}$

$\textsf{Here,} \\ \bullet\:\textsf{The value of y = sin(ax + b)} \\ \bullet\:\textsf{The value of u = (ax + b)}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[sin(ax + b)]}{d(ax + b)}\cdot\dfrac{d(ax + b)}{dx}}$

$\textsf{Now, by using the formulas and substituting them in the equation, we get :-} \\ \\ \bullet\:\underline{\sf{Derivative\:of\:sin(x)\:is \:cos(x)}} \\ \sf{\dfrac{d[sin(x)]}{dx} = cos(x)} \\ \\ \underline{\sf{Sum\:rule\: differentiation :-}} \\ \\ \sf{\dfrac{d}{dx}[f(x) + g(x)] = \dfrac{d[f(x)]}{du} + \dfrac{d[g(x)]}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = cos(ax + b)\cdot\bigg[\dfrac{d(ax)}{dx} + \dfrac{d(b)}{dx}\bigg]}$

$\textsf{By applying the constant rule of differentiation and substituting the values in it,}\\\textsf{we get :-}$

$\underline{\sf{Constant\:rule\: differentiation :-}} \\ \\ \sf{\dfrac{dc}{dx} = 0}$

$:\implies \sf{\dfrac{dy}{dx} = cos(ax + b)\cdot(a + 0)}$

$:\implies \sf{\dfrac{dy}{dx} = cos(ax + b)a}$

$\boxed{\therefore \sf{\dfrac{dy}{dx} = cos(ax + b)a}}$

$\textsf{Thus, the derivative of sin(ax + b) is cos(ax + b)a.}$

To find the second derivative of the function sin(ax + b) :

$\sf{Function\:of\:y = cos(ax + b)a}$

$\textsf{By applying the product rule of differentiation and substituting the values in it,}\\\textsf{we get :-}$

$\underline{\sf{Product\:rule\: differentiation :-}} \\ \\ \sf{\dfrac{dy}{dx} = (u)\cdot\dfrac{d(v)}{dx} + (v)\cdot\dfrac{d(u)}{dx}}$

$\textsf{Here,} \\ \bullet\:\textsf{The value of u = cos(ax + b)} \\ \bullet\:\textsf{The value of v = a}$

$:\implies \sf{\dfrac{dy}{dx} = cos(ax + b)\cdot\dfrac{d(a)}{dx} + (a)\cdot\dfrac{d[cos(ax + b)]}{dx}}$

$\textsf{By applying the constant rule of differentiation and substituting the values in it,}\\\textsf{we get :-}$

$\underline{\sf{Constant\:rule\: differentiation :-}} \\ \\ \sf{\dfrac{dc}{dx} = 0}$

$:\implies \sf{\dfrac{dy}{dx} = cos(ax + b)\cdot(0) + (a)\cdot\dfrac{d[cos(ax + b)]}{dx}}$

$\textsf{Now, By applying the chain rule of differentiation and substituting the values in it,}\\\textsf{we get :-}$

$\underline{\sf{Chain\:rule\: differentiation :-}} \\ \\ \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}}$

$\textsf{Here,} \\ \bullet\:\textsf{The value of y = cos(ax + b)} \\ \bullet\:\textsf{The value of u = (ax + b)}$

$:\implies \sf{\dfrac{dy}{dx} = (a)\cdot\bigg[\dfrac{d[cos(ax + b)}{d(ax + b)}\cdot\dfrac{d(ax + b)}{dx}\bigg]}$

$\textsf{Now, by using the formulas and substituting them in the equation, we get :-} \\ \\ \bullet\:\underline{\sf{Derivative\:of\:cos(x)\:is \:-sin(x)}} \\ \sf{\dfrac{d[cos(x)]}{dx} = -sin(x)} \\ \\ \underline{\sf{Sum\:rule\: differentiation :-}} \\ \\ \sf{\dfrac{d}{dx}[f(x) + g(x)] = \dfrac{d[f(x)]}{du} + \dfrac{d[g(x)]}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = (a)\cdot\bigg[-sin(ax + b)\cdot\dfrac{d(ax)}{dx} + \dfrac{d(b)}{dx}\bigg]}$

$\textsf{By applying the constant rule of differentiation and substituting the values in it,}\\\textsf{we get :-}$

$\underline{\sf{Constant\:rule\: differentiation :-}} \\ \\ \sf{\dfrac{dc}{dx} = 0}$

$:\implies \sf{\dfrac{dy}{dx} = (a)\cdot[-sin(ax + b)\cdot(a + 0)]}$

$:\implies \sf{\dfrac{dy}{dx} = -sin(ax + b)a^{2}}$

$\boxed{\therefore \sf{\dfrac{dy}{dx} = -sin(ax + b)a^{2}}}$

$\sf{Thus,\:the\:derivative\:of\:cos(ax + b)a\:is\:-sin(ax + b)a^{2}.}$

Answer 2

Solution

Let y = sin ( ax + b )

$\sf \: then \: \: \: \dfrac{dy}{dx} = \: \dfrac{d}{dx} \: \sin(ax + b)$

$\sf \implies \: \cos( \: ax + b \: ) \: \dfrac{d}{dx} \: (ax \: + \: b)$

$\sf \implies \: \cos( \: ax + \: b \: ) \times ( \: a \: x \: 1 + b \: )$

$\implies \sf \: a \sin \dfrac{\pi}{2} + ( \: ax + b \: )$

$\sf \dfrac{ {d}^{2}y }{ {dx}^{2} } \: = \dfrac{d}{dx} \: a \: \cos( \: ax + b \: )$

$\sf \implies \: a \dfrac{d}{dx} \cos( \: ax + b \: )$

$\sf \implies \: a( - \sin( \: ax + b \: ) ). \: \dfrac{d}{dx} ( \: ax + b \: )$

$\implies \sf \: a( - \: sin \: ( \: ax + b \: )) \times ( \: a \: x \: 1 + 0 \: )$

$\implies \sf {a}^{2} .sin(\pi + ( \: ax + b \: ))$

$\sf \implies \: {a}^{2} .sin \: ( \dfrac{2\pi}{2} + ( \: ax + b \: ))$

$\sf \dfrac{ {d}^{3}y }{d {x}^{3} } = \dfrac{d}{dx} ( - {a}^{2} \: \sin( \: x + b \: )$

$\sf \implies \: - {a}^{2} \dfrac{d}{dx} \: sin( \: ax + b \: )$

$\sf \implies \: - {a}^{2} . \cos( \: ax + b \: ) . \: \dfrac{d}{dx}( \: ax + b \: )$

$\sf \implies \: - {a}^{2} . \cos( \: ax + b \: ) \times (a \: x \: 1 + 0)$

$\sf \implies \: {a}^{3} . \sin \dfrac{3\pi}{2} + ( \: ax + b \: )$

In general , the nth order derivative is given by

$\sf \dfrac{{d}^{n}y}{d {x}^{n} } = {a}^{n} . \sin \dfrac{n\pi}{2} + ( \: ax + b \: )$