Question

Derivative of sin(ax+b)cos(cx+d) with respect to x is

Answer 1

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

Now we will use product rule :-

$\underline{ \boxed{ \large\sf\dfrac{d(uv)}{dx} = u \dfrac{dv}{dx} + v \dfrac{du}{dx}}}$

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

We know that :-

$\underline{ \boxed{ \large\sf\dfrac{d(cos \: t)}{dt} = - sin \: t}}$

$\underline{ \boxed{ \large\sf\dfrac{d(sin \: t)}{dt} = cos \: t }}$

$\underline{ \boxed{ \large\sf\dfrac{d( {t}^{m} )}{dt} = m \: {t}^{m - 1} }}$

$\sf \implies - sin \: (ax + b) sin(cx + d)(c + 0)+ cos(cx + d) cos(ax + b)(a + 0)$

$\sf \implies - c \: sin \: (ax + b) sin(cx + d) +a \: cos(cx + d) cos(ax + b)$

Answer :

$\sf \dfrac{d[sin \: (ax + b) \: cos(cx + d)]}{dx} = - c \: sin \: (ax + b) sin(cx + d) +a \: cos(cx + d) cos(ax + b)$

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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