How do you find the derivatives of #y=(sintheta)^tantheta# by logarithmic differentiation?

Question

William Abdalla · Accepted Answer

#(dy)/(dx)=(sintheta)^(tantheta)(sec^2thetalnsintheta+1)#

usinf logarithmic differentiation

#y=(sintheta)^(tantheta)#

#lny=ln(sintheta)^(tantheta)#

by laws of logs

#lny=tanthetaln(sintheta)#

now differentiate #wrt" " x#

#d/dx(lny=tanthetaln(sintheta))#

#RHS" "# will need the product rule

#1/y(dy)/(dx)=sec^2thetalnsintheta+tantheta1/sinthetaxxcostheta#

#1/y(dy)/(dx)=sec^2thetalnsintheta+cancel((sintheta/costhetaxxcostheta/sintheta))^(=1)#

#1/y(dy)/(dx)=sec^2thetalnsintheta+1#

#(dy)/(dx)=y(sec^2thetalnsintheta+1)#

#(dy)/(dx)=(sintheta)^(tantheta)(sec^2thetalnsintheta+1)#

William Abdalla · Answer

undefined To find the derivative of $ y = (\sin(\theta))^{\tan(\theta)} $ using logarithmic differentiation:

1. Take the natural logarithm of both sides: $ \ln(y) = \ln\left((\sin(\theta))^{\tan(\theta)}\right) $.
2. Use the properties of logarithms to simplify the expression: $ \ln(y) = \tan(\theta) \cdot \ln(\sin(\theta)) $.
3. Differentiate both sides with respect to $ \theta $.
4. Apply the chain rule and product rule as necessary.
5. Solve for $ \frac{dy}{d\theta} $ to find the derivative.

The derivative can be expressed as $ \frac{dy}{d\theta} = \frac{\cos(\theta) \ln(\sin(\theta)) + \frac{\tan(\theta)}{\sin(\theta)}}{(\sin(\theta))^{\tan(\theta)}} $.