How do you use logarithmic differentiation to find the derivative of #y=(cosx)^x#?

Question

Elijah Abram · Accepted Answer

#(dy)/(dx)=(cosx)^x[lncosx-xtanx]#

#y=(cosx)^x#

take natural logs of bothe sides

#lny=ln(cosx)^x#

#=>lny=xlncosx#

now differentite #wrt" "x#

the #RHS will need the product rule

#d/(dx)(lny=xlncosx)#

#=>d/(dx)(lny)=lncosxd/(dx)(x)+xd/(dx)(lncosx)#

#1/y(dy)/(dx)=lncosx+x xx (-sinx)/(cosx)#

#(dy)/(dx)=y[lncosx-xtanx]#

substitute for #y#

#(dy)/(dx)=(cosx)^x[lncosx-xtanx]#

Aurora Alvarado · Answer

undefined To find the derivative of $ y = (\cos(x))^x $ using logarithmic differentiation, follow these steps:

1. Take the natural logarithm of both sides of the equation: 
$$ \ln(y) = \ln\left((\cos(x))^x\right) $$

2. Use the properties of logarithms to simplify the expression: 
$$ \ln(y) = x \ln(\cos(x)) $$

3. Differentiate both sides of the equation with respect to $ x $: 
$$ \frac{d}{dx}(\ln(y)) = \frac{d}{dx}(x \ln(\cos(x))) $$

4. Apply the chain rule and product rule on the right-hand side: 
$$ \frac{1}{y} \frac{dy}{dx} = \ln(\cos(x)) + x \left(\frac{-\sin(x)}{\cos(x)}\right) $$

5. Solve for $ \frac{dy}{dx} $: 
$$ \frac{dy}{dx} = y \left(\ln(\cos(x)) - \frac{x\sin(x)}{\cos(x)}\right) $$

6. Substitute back the original expression for $ y $: 
$$ \frac{dy}{dx} = (\cos(x))^x \left(\ln(\cos(x)) - \frac{x\sin(x)}{\cos(x)}\right) $$

So, the derivative of $ y = (\cos(x))^x $ with respect to $ x $ using logarithmic differentiation is:
$$ \frac{dy}{dx} = (\cos(x))^x \left(\ln(\cos(x)) - \frac{x\sin(x)}{\cos(x)}\right) $$