Use Logaritmic Differentiation to Find the Derivative

[alexmx2007]

I think I did something wrong here:

Using Logarithmic Differentiation:

y=(sinxcosxtan^3x)/square root x

lny=lnsinx+lncosx+lntan^3x-ln(x)^1/2

lny=1/sinx * cosx + 1/cosx * sinx + 1/tan^3x - 1/2ln(x) I THINK THIS IS WHERE I MESSED UP SOMEHOW.

1/y dy/dx= cot(x)+tan(X) + 1/tan^3x-1/2ln(x)

(Y)1/y dy/dx= (cot(x) + 1/tan^2x-1/2x)(Y)

dy/dx= (cotx+1/tan^2x-1/2x)((sinxcosxtan^3x)/square root x) IS THIS RIGHT?

 

[Deleted member 4993]

I think I did something wrong here:

Using Logarithmic Differentiation:

y=(sinxcosxtan^3x)/square root x

lny=lnsinx+lncosx+lntan^3x-ln(x)^1/2

lny=1/sinx * cosx + 1/cosx * sinx + 1/tan^3x - 1/2ln(x) I THINK THIS IS WHERE I MESSED UP SOMEHOW.

1/y dy/dx= cot(x)+tan(X) + 1/tan^3x-1/2ln(x)

(Y)1/y dy/dx= (cot(x) + 1/tan^2x-1/2x)(Y)

dy/dx= (cotx+1/tan^2x-1/2x)((sinxcosxtan^3x)/square root x) IS THIS RIGHT?

Why areyou trying use "log" here? Why not straight-forward use of productrule (and chain rule and quotient rule)?

 

[alexmx2007]

I used rules of lograithims because the question says to find the derivative by using logarithmic differientation.

 

[BigGlenntheHeavy]

$\displaystyle I'll \ get \ you \ started:$

$\displaystyle y \ = \ \frac{sin(x)cos(x)tan^3(x)}{\sqrt x}, \ ln|y| \ = \ ln \ \bigg|\frac{sin(x)cos(x)tan^3(x)}{\sqrt x}\bigg|.$

$\displaystyle ln|y| \ = \ ln|sin(x)|+ln|cos(x)|+3ln|tan(x)|-\frac{ln|x|}{2}.$

$\displaystyle \frac{y'}{y} \ = \ cot(x)-tan(x)+\frac{3}{sin(x)cos(x)}-\frac{1}{2x}.$

$\displaystyle y' \ = \ \frac{sin(x)cos(x)tan^3(x)}{x^{1/2}}\bigg[\frac{2xcos^2(x)-2xsin^2(x)+6x-sin(x)cos(x)}{2xsin(x)cos(x)}\bigg].$

$\displaystyle Now, \ I'll \ let \ you \ have \ the \ rest \ of \ the \ fun.$