Your question basically asks, #d/dx(e^sqrt(x)cos(x))/(cos(x)-sin(x))#, when #x=pi/3#.
First, we must solve for the derivative. Apply the quotient rule:
#(f/g)'=(f'g-fg')/g^2#
#(d/dx(e^sqrt(x)cos(x))(cos(x)-sin(x))-d/dx(cos(x)-sin(x))(e^sqrt(x)cos(x)))/(cos(x)-sin(x))^2#
We must calculate #d/dxe^sqrt(x)cos(x)#.
Apply the product rule:
#(f*g)'=f'g+fg'#
The derivative of #e^sqrt(x)# is:
Apply the chain rule: #(df(u))/dx=(df)/(du)*(du)/dx#. Here, #e^u=f# and #sqrt(x)=u#
The chain rule calculations spit out #(e^sqrt(x))/(2sqrt(x))#, which is the derivative of #e^sqrt(x)#.
The derivative of #cos(x)# is #-sin(x)#.
#d/dxe^sqrt(x)cos(x)=(e^sqrt(x)cos(x))/(2sqrt(x))-e^sqrt(x)sin(x)#.
Now we must calculate #d/dxcos(x)-sin(x)#
Use the difference rule:
#(f-g)'=f'-g'#
The derivative of #sin(x)# is #cos(x)#. The derivative of #cos(x)# is #-sin(x)#
#d/dxcos(x)-sin(x)=-sin(x)-cos(x)#
Now input that into our very first derivative, the one using the quotient rule:
#((e^sqrt(x)cos(x)))/(2sqrt(x))-e^sqrt(x)sin(x)(cos(x)-sin(x)-(-sin(x)-cos(x)*(e^sqrt(x)cos(x))))/(cos(x)-sin(x))^2#
This simplifies into:
#(2e^sqrt(x)-e^sqrt(x)sin(2x)+4e^sqrt(x)sqrt(x)cos^2(x)+4e^sqrt(x)sqrt(x)sin^2(x))/(4sqrt(x)(cos(x)-sin(x))^2)#.
This is the equation of the slope at any #x#.
When #x=pi/3#, the slope is:
#(2e^sqrt(pi/3)-e^sqrt(pi/3)sin(2*pi/3)+4e^sqrt(pi/3)sqrt(pi/3)cos^2(pi/3)+4e^sqrt(pi/3)sqrt(pi/3)sin^2(pi/3))/(4sqrt(pi/3)(cos(pi/3)-sin(pi/3))^2)#
Which simplifies to #26.52188#.
