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ekw606
ekw606 ekw606
  • 23-07-2016
  • Mathematics
Answered

I'm not sure where to start and how to differentiate this.

I m not sure where to start and how to differentiate this. class=

Verified Answer :

padawan
padawan padawan
  • 23-07-2016

the picture is not clear. let me assume
y = (x^4)ln(x^3)


product rule :

d f(x)g(x) = f(x) dg(x) + g(x) df(x)


dy/dx = (x^4)d[ln(x^3)/dx] + d[(x^4)/dx] ln(x^3)

= (x^4)d[ln(x^3)/dx] + 4(x^3) ln(x^3)


look at d[ln(x^3)/dx]

d[ln(x^3)/dx]

= d[ln(x^3)/dx][d(x^3)/d(x^3)]

= d[ln(x^3)/d(x^3)][d(x^3)/dx]

= [1/(x^3)][3x^2] = 3/x

... chain rule (in detail)


end up with

dy/dx = (x^4)[3/x] + 4(x^3) ln(x^3)

= x^3[3 + 4ln(x^3)]

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