Thread: Implicit Differentiation Question

Hi

Could someone please show me how to differentiate y = [ln(x)]^(sinx)?

Thanxx a lot!

y = ln(x)^(sin(x))

ln(y) = sin(x)ln(ln(x))

1/y dy/dx = cos(x)ln(ln(x) + sin(x) (1/ln(x))1/x

dy/dx = y[cos(x)ln(ln(x) + sin(x) (1/ln(x))1/x]

dy/dx = ln(x)^(sin(x))[cos(x)ln(ln(x) + sin(x) (1/ln(x))1/x]

Originally Posted by Calculus26

y = ln(x)^(sin(x))

ln(y) = sin(x)ln(ln(x))

1/y dy/dx = cos(x)ln(ln(x) + sin(x) (1/ln(x))1/x

dy/dx = y[cos(x)ln(ln(x) + sin(x) (1/ln(x))1/x]

dy/dx = ln(x)^(sin(x))[cos(x)ln(ln(x) + sin(x) (1/ln(x))1/x]

Can that part be simplified to:

sin(x) / (x ln(x))
= 1 / ln(x)

How do you get sin(x)/[xln(x)] simplifying to 1/ln(x) ?

you are not suggesting sin(x)/x =1 are you?

Originally Posted by xwrathbringerx

Can that part be simplified to:

sin(x) / (x ln(x))
= 1 / ln(x)

but, in general, .