Question

D/dx(sinx+cosx) is equal to

Answer 1

$\implies \: \dfrac{d( \sin \: x+ \cos \: x) }{dx}$

$\implies \: \dfrac{d( \sin \: x) }{dx} + \dfrac{d( \cos \: x) }{dx}$

We know that :-

$\dfrac{d( \sin \: t) }{dt} = \cos \: t$

$\dfrac{d( \cos \: t)}{dt} = - \sin \: t$

$\implies \: {( \cos \: x) } + {( - \sin \: x) }$

$\implies \: {\cos \: x} - \sin \: x$

Answer :

$\dfrac{d( \sin \: x+ \cos \: x) }{dx} = {\cos \: x} - \sin \: x$

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Learn more :

d(x^n)/dx = n x^(n-1)

d(log x)/dx = 1/x

d(e^x)/dx = e^x

d(sinx)/dx = cosx

d(cos x)/dx = -sin x

d(cosec x)/dx = -cot x cosec x

d(tan x)/dx = sec²x

d(sec x)/dx = secx tanx

d(cot x)/dx = - cosec² x