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Answer:
dydx=3sin2(x)⋅cosx
Explanation:
In order to differentiate sin3(x), we need to use a chain rule, which tells us that
ddx[f(g(x))]=f'(g(x))⋅g'(x)
Letting y=sin3(x), then
dydx=3sin2(x)⋅cosx
In this problem, we've also performed the power rule, namely by subtracting 1 from the power of 3 on the sinx term, which is why we end up with a sin2(x).
dydx=3sin2(x)⋅cosx
Explanation:
In order to differentiate sin3(x), we need to use a chain rule, which tells us that
ddx[f(g(x))]=f'(g(x))⋅g'(x)
Letting y=sin3(x), then
dydx=3sin2(x)⋅cosx
In this problem, we've also performed the power rule, namely by subtracting 1 from the power of 3 on the sinx term, which is why we end up with a sin2(x).
Answered by
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let Cos^3 x = t
- 3 cos^2 x sinx dx = dt
let sin^3x = a
3 sin^2x cos x dx = da
da/dt = 3 sin^2 x cosx / -3 cos^2x sin x
= - tan^2x cot x
= - tan x
- 3 cos^2 x sinx dx = dt
let sin^3x = a
3 sin^2x cos x dx = da
da/dt = 3 sin^2 x cosx / -3 cos^2x sin x
= - tan^2x cot x
= - tan x
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