find the derivative

Question

find the derivative:

let y(x)=(10(sin x)^-7)/x

what would the first derivative be?

Answer 1

y = 10(sinx)^-7/x = 10(sinx)^-7(x^-1)

use the product rule or the quotient rule

what follows 1st is the product rule

derivative of uv = uv'+vu' u=10sinx^-7, v=x^-1, u'=-70sin^-8cosx, v'=-x^-2

derivative = 1st term times derivative of 2nd term + 2nd term times derivative of the 1st term

y'= 10(sinx)^-7(-1/x^2) + (1/x)(-70)(sinx)^-8(cosx)

y'= = -10/(x^2)(sinx)^7 -70cosx/xsinx^8

y'= (-10sinx - 70xcosx))/x^2sinx^8

y'= -(70xcosx +10sinx)/(x^2)(sinx^8

or use the quotient rule

derivative = denominator times derivative of numerator minus numerator times derivative of denominator, all over denominator squared

(u/v)'= (vu' - uv')/v^2

y = 10/xsinx^7

10=u, xsinx^7 = v

u'=0, v' = 7xcosxsinx^6 +sinx^7

y' = (vu'-uv')/v^2 = (0 - 10[(x)(7sin^6)cosx + sinx^7]/x^2sinx^14

divide numerator and denominator by sinx^6

y' = -(70xcosx +10sinx)/x^2sinx^8

it gets complicated or tedious enough, it's easy to make a mistake in signs or arithmetic