Question

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) =

Answer 1

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

Given that,

☆ f(x) is derivative of g(x)

$\bf\implies \:f(x) = g'(x) - - - (1)$

Again,

Given that

☆ g(x) is the derivative of h(x)

$\bf\implies \:g(x) = h'(x) - - - (2)$

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

$\bf\implies \:f(x) = h''(x) - - - (3)$

Now,

Given that,

$\rm :\longmapsto\:h(x) = a \: sinx \: + \: b \: cosx \: + \: c$

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

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

$\because \green{\boxed{ \bf{ \:\dfrac{d}{dx}sinx = cosx}}} \: and \: \green{\boxed{ \bf{ \:\dfrac{d}{dx}cosx = - sinx}}}$

$\rm :\longmapsto\:h'(x) = a \: cosx \: - \: b \: sinx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}h'(x) = \dfrac{d}{dx}(a \: cosx \: - \: b \: sinx)$

$\bf :\longmapsto\:h''(x) = - \: a \: sinx \: - \: b \: cosx - - - (4)$

Now,

Consider,

$\bf :\longmapsto\:f(x) + h(x)$

$\rm \: \: = \: h''(x) \: + \: h(x)$

$\: \: \: \: \green{\boxed{ \bf{ \: \because \: using \: (3)}}}$

$\rm \: \: = \: - asinx \: - \: bcosx \: + \: asinx \: + \: bcosx \: + \: c$

$\rm \: \: = \: c$

Hence,

$\bf :\longmapsto\:f(x) + h(x) = c$