How Do You Differentiate y = sin(x)^x Using Natural Logs?

gyza502
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Homework Statement


Let y=sin(x)^x. Find dy/dx


My teacher said to use Natural long to solve this.

she got,
step 1:ln(y)=ln(sin(x)^x)=x ln(sin(x))

step 2:y(dy/dx)=1(ln(sin(x)+x(cos(x)/sin(x)

Final answer:dy/dx=((sin(x))^x)[ln(sin(x))+x(cos x/sin x)

I don't understand how she derived ln sin(x) to be cos(x)/sin(x)...
 
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gyza502 said:

Homework Statement


Let y=sin(x)^x. Find dy/dx


My teacher said to use Natural long to solve this.

she got,
step 1:ln(y)=ln(sin(x)^x)=x ln(sin(x))

step 2:y(dy/dx)=1(ln(sin(x)+x(cos(x)/sin(x)

Final answer:dy/dx=((sin(x))^x)[ln(sin(x))+x(cos x/sin x)

I don't understand how she derived ln sin(x) to be cos(x)/sin(x)...
Hello gyza502. Welcome to PF !

Use the derivative of the natural log, along with the chain rule.
 
Let u= sin(x) so that y= ln(u)

By the chain rule, dy/dx= (dy/du)(du/dx).

What is dy/du? What is du/dx? Be sure to express every thing in terms of x, not u.
 
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