y = (x-4)^-3(sin(x))^4 / (x^2-2x)^2(e^x^3)

y'= y*____________

I set it up as:

sin^4(x) / (x-4)^3(x^2-2x)^2(e^x^3)

Take the ln of both sides:

ln y= 4sin^3(x)cos(x) / ((3(x-4)^2)(2(x^2-2x))(2x-1)(3x))

But that's wrong. What do I do with this?

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- Feb 3rd 2010, 06:57 PMoperaphantom2003Use logarithmic differentiation to find dy/dx, where
y = (x-4)^-3(sin(x))^4 / (x^2-2x)^2(e^x^3)

y'= y*____________

I set it up as:

sin^4(x) / (x-4)^3(x^2-2x)^2(e^x^3)

Take the ln of both sides:

ln y= 4sin^3(x)cos(x) / ((3(x-4)^2)(2(x^2-2x))(2x-1)(3x))

But that's wrong. What do I do with this? - Feb 3rd 2010, 07:14 PMCalculus26
Use ln(a/b) = lna - lnb

and ln(a^b) = bln(a)

y = (x-4)^-3(sin(x))^4 / (x^2-2x)^2(e^x^3)

lny = [-3sin(x)] ln(x-4) -[ e^x^3] ln (x^2 -2x)

Now differentiate:

y '/y = etc....