Hi Could someone please show me how to differentiate y = [ln(x)]^(sinx)? Thanxx a lot!
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$\displaystyle (\ln x)^{\sin x}=e^{\ln \ln x\sin x}$
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) $\displaystyle \lim_{x\rightarrow 0} \frac{sin(x)}{x}= 1$ but, in general, $\displaystyle \frac{sin(x)}{x}\ne 1$.
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