Suppose f(x) is the derivative of g(x) and g(x)is the derivative of h(x).If h(x) = asin x + bcos x + c then f(x) + h(x) =

1 thought on “Suppose f(x) is the derivative of g(x) and g(x)is the derivative of h(x).If h(x) = asin x + bcos x + c then f(x) + h(x) =”

1. [tex]\large\underline{\sf{Solution-}}[/tex]

   Given that,

   ☆ f(x) is derivative of g(x)

   [tex]\bf\implies \:f(x) = g'(x) – – – (1)[/tex]

   Again,

   Given that

   ☆ g(x) is the derivative of h(x)

   [tex]\bf\implies \:g(x) = h'(x) – – – (2)[/tex]

   ☆ Substituting equation (2) in equation (1), we get

   [tex]\bf\implies \:f(x) = h”(x) – – – (3)[/tex]

   Now,

   Given that,

   [tex]\rm :\longmapsto\:h(x) = a \: sinx \: + \: b \: cosx \: + \: c[/tex]

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

   [tex]\rm :\longmapsto\:\dfrac{d}{dx}h(x) = \dfrac{d}{dx}(a \: sinx \: + \: b \: cosx \: + \: c)[/tex]

   [tex] \because \green{\boxed{ \bf{ \:\dfrac{d}{dx}sinx = cosx}}} \: and \: \green{\boxed{ \bf{ \:\dfrac{d}{dx}cosx = – sinx}}}[/tex]

   [tex]\rm :\longmapsto\:h'(x) = a \: cosx \: – \: b \: sinx[/tex]

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

   [tex]\rm :\longmapsto\:\dfrac{d}{dx}h'(x) = \dfrac{d}{dx}(a \: cosx \: – \: b \: sinx)[/tex]

   [tex]\bf :\longmapsto\:h”(x) = – \: a \: sinx \: – \: b \: cosx – – – (4)[/tex]

   Now,

   Consider,

   [tex]\bf :\longmapsto\:f(x) + h(x)[/tex]

   [tex] \rm \: \: = \: h”(x) \: + \: h(x)[/tex]

   [tex] \: \: \: \: \green{\boxed{ \bf{ \: \because \: using \: (3)}}}[/tex]

   [tex] \rm \: \: = \: – asinx \: – \: bcosx \: + \: asinx \: + \: bcosx \: + \: c[/tex]

   [tex] \rm \: \: = \: c[/tex]

   Hence,

   [tex]\bf :\longmapsto\:f(x) + h(x) = c[/tex]

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