Slope of The Tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$ Find slope of tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$
using $$\frac{dy}{dx} = \frac{\frac{dr}{dθ}\sin \theta + r \cos \theta}{\frac{dr}{dθ}\cos \theta-r \sin \theta}$$
I got
$$\frac {14\cos \theta \sin \theta}{7 \cos^2 \theta-7\sin \theta cos \theta}$$
 A: The curve has polar equation
$r=7 \sin t$ 
multiply both sides by $r$
$r^2=7r\sin t$
in rectangular coordinates becomes
$x^2+y^2=7y$
Equation of a circle having center $C(0,\;7/2)$ and radius $R=7/2$
The point of the circle when $t=\pi/6$ can be found intersecting the line, in polar form, $a:t=\pi/6$ with the given circle. The line $a$ in rectangular coordinates has equation 
$$y=x\tan(\pi/6)\to x=y\sqrt 3$$
plugging this in the equation of the circle we get
$3y^2+y^2=7y\to y=0;\;y=7/4$
Therefore coordinates of $P$ are $(7/4\sqrt 3,\;7/4$)
To find the slope of the tangent in $P$ we write the equation of the polar line to the circle in $P$. The polar line of a conic in a point lying on the conic is the tangent to the conic in that point.
To find the polar line substitute in the equation of the conic$^{(*)}$
$x^2\to x_Px;\;y^2\to y_Py;\;y\to \dfrac{y+y_P}{2}$
and get
$\dfrac{7}{4}x\sqrt 3+\dfrac{7}{4}y =7\,\dfrac{y+\frac{7}{4}}{2}$
$x\sqrt 3+y=2\left(y+\frac{7}{4}\right)$
$y=x\sqrt{3} -\dfrac{7}{2}$
and finally the slope $\color{red}{m=\sqrt 3}$
Ehm... maybe not so much simpler :)
$(*)$ to complete the formulas
$x\to \dfrac{x+x_p}{2};\;xy\to \dfrac{x_Py+y_Px}{2}$
$$...$$
Hope this can be useful
A: When you have the equation in rectangular coordinates
$$
x^2+y^2=7y
$$
you can use implicit differentiation:
$$
2x+2yy'=7y'
$$
so
$$
y'=\frac{2x}{7-2y}
$$
For $\theta=\pi/6$, we have $r=7/2$, so $x=7\sqrt{3}/4$ and $y=7/4$; therefore the derivative at the point is
$$
y'(7\sqrt{3}/4)=\frac{2\dfrac{7\sqrt{3}}{4}}{7-2\dfrac{7}{4}}=\sqrt{3}
$$
If you want to do it directly with polar coordinates,
$$
dx=\cos\theta\,dr-r\sin\theta\,d\theta
\qquad
dy=\sin\theta\,dr+r\cos\theta\,d\theta
$$
which gives
$$
\frac{dy}{dx}
=\frac{\sin\theta\,dr+r\cos\theta\,d\theta}{\cos\theta\,dr-r\sin\theta\,d\theta}
=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}
=\tan2\theta
$$
Your derivation is almost correct: you have $\sin\theta\cos\theta$ at the bottom right, but it should be $7\sin^2\theta$.
A: Your derivation seems to be not correct.
Indeed:
$$x=r \cos \theta$$
$$y=r \sin \theta$$
$$r=7 \sin \theta \implies dr=7 \cos \theta d\theta$$
Thus:
$$dx=\cos \theta dr - r \sin \theta d\theta=7\cos^2 \theta-7\sin^2 \theta$$
$$dy=\sin \theta dr +r \cos \theta d\theta=7\cos \theta \sin \theta+7\cos \theta \sin \theta$$
That is
$$\frac{dy}{dx}=\frac {2\cos \theta \sin \theta}{\cos^2 \theta-\sin^2 \theta}=\frac{\frac{\sqrt 3}{2}}{\frac{3}{4}-\frac{1}{4}}=\sqrt 3$$
