Question

What is the nth derivatives of
Y= sin³xcos²x
Please help

Answer 1

Let y=sin2x⋅cos3x=(1−cos2x2)⋅(cos3x+3cosx4)

=18⋅(cos3x+3cosx−cos2xcos3x−3cos2xcosx)

=18⋅(cos3x+3cosx−cos(3x−2x)+cos(3x+2x)2−3(cos(2x−x)+cos(2x+x))2)

=116⋅(2cos3x+6cosx−cosx−cos5x−3cosx−3cos3x)

⟹y=116⋅(2cosx−cos3x−cos5x)

From : [math]n^{th} \text{[/math]

Footnotes

If you know your trig identities, then you know that sin2x=1−cos2x , which you can substitute into the expression you have to get u=sin2xcos3x=(1−cos2x)cos3x=cos3x−cos5x .

Let’s do a substitution, of c=cosx , so we are now trying to find the derivative u′=(3c2−5c4)c′ . If you want to turn that back into terms of sinx,cosx , simply reverse the substitution, keeping in mind that c′=−sinx= . If we call s=−c′ , we can express this as −s(3c3−c5) , which only superficially seems better. But keeping in mind that s′′=−s,c′′=−c , as well as s2+c2=1 , we can potentially do a bit of simplification as we go.

So we have u′=−s(3c2−5c4) . Using our identities, we can get u′=−s(3(1−s2)−5(1−s2)2)=−s(3–3s2−5(1–2s2+s4))=−3s+3s3+5s−10s3+5s5=5s5−7s3+2s .

So the original function is of the form u=A0c5+B0c3+C0c , and the first derivative is of the form u′=A1s5+b1s3+C1s .

Is that going to continue? Let’s see.

v=Ac5+Bc3+Cc

v′=−s(5Ac4+3Bc2+C)

v′=−s(5A(1−s2)2+3B(1−s2)+C)

v′=−s(5A(1–2s2+s4)+3B−3Bs2+C)

v′=−5As5+(10A+3B)s3−(5A+3B+C)s

w=As5+Bs3+Cs

w′=c(5As4+3Bs2+C)

w′=c(5A(1−c2)2+3B(1−c2)+C)

w′=c(5A(1−2c2+c4)+3B−3Bc2+C)

w′=c(5Ac4−(10A+3B)c2+3B+C+5A)

w′=5Ac5+(−10A−3B)c3+(5A+3B+C)c

So if you have a quintic in sines with no even terms, the derivatives is a quintic in cosines with no even terms, and vice-versa.

So all you have to do is compute the three coefficients, which all depend on each other. If you follow my derivation correctly (i.e., excluding any errors I might have inadvertently inserted), you can get expressions for the coefficients of u(n) in terms of the coefficients of u(n−1) .

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Answer 2

Answer:

Step-by-step explanation: