Question

the nth derivative of 5xpower n​

Answer 1

Step-by-step explanation:

Remember that:

cosh ix = cosx → cosh x = cos(x/i) = cos(-ix) = cos(ix)

so

y= cosh4xcos3x

y= cos(4ix)cos3x

y= (1/2)(cos(3+4i)x + cos(3–4i)x)

y´= (1/2)(-(3+4i)sin(3+4i)x -(3–4i)sin(3–4i)x)

y´= (-1/2)((3+4i)sin(3+4i)x+(3–4i)sin(3–4i)x)

y´´= (-1/2)((3+4i)^2cos(3+4i)x+(3–4i)^2cos(3–4i)x)

y´´´= (+1/2)((3+4i)^3sin(3+4i)x+(3–4i)^3sin(3–4i)x)

y(4)´= (+1/2)((3+4i)^4cos(3+4i)x+(3–4i)^4cos(3–4i)x)

y(5)´= (-1/2)((3+4i)^5sin(3+4i)x+(3–4i)^5sin(3–4i)x)

$\huge\color{green}\boxed{\colorbox{black}{PLS FOLLOW}}$

Answer 2

Step-by-step explanation:

Remember that:

cosh ix = cosx → cosh x = cos(x/i) = cos(-ix) = cos(ix)

so

y= cosh4xcos3x

y= cos(4ix)cos3x

y= (1/2)(cos(3+4i)x + cos(3–4i)x)

y´= (1/2)(-(3+4i)sin(3+4i)x -(3–4i)sin(3–4i)x)

y´= (-1/2)((3+4i)sin(3+4i)x+(3–4i)sin(3–4i)x)

y´´= (-1/2)((3+4i)^2cos(3+4i)x+(3–4i)^2cos(3–4i)x)

y´´´= (+1/2)((3+4i)^3sin(3+4i)x+(3–4i)^3sin(3–4i)x)

y(4)´= (+1/2)((3+4i)^4cos(3+4i)x+(3–4i)^4cos(3–4i)x)

y(5)´= (-1/2)((3+4i)^5sin(3+4i)x+(3–4i)^5sin(3–4i)x)