Note: This is part 3 rd of a 4 part question, split as 4 separate questions here.

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- If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
- $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
- The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$

$y=2 \sin ^2 3x$ at $x=\large\frac{\pi}{6}$

Step 1:

When $x=\large\frac{\pi}{6},$$y=2 \sin ^2 \large\frac{\pi}{2}$$=2$

We are required to find the equation of the tangent and normal at $\bigg(\large\frac{\pi}{6}$$,2 \bigg)$

Step 2:

Slope of the tangent at $\bigg(\large\frac{\pi}{6},$$2\bigg)=\large\frac{dy}{dx_{(\Large\frac{\pi}{6},2)}}$

$\large\frac{dy}{dx}$$=4 \sin 3x. 3 \cos 3x=12 \sin 3x \cos 3x=6 \sin 6x$

$m=\large\frac{dy}{dx_{(\Large\frac{\pi}{6},2)}}=$$6 \sin \large\frac{\pi}{n}=0$

Therefore the tangent is $||$ to the x axes and to equation is $y=2$

Step 3:

The normal is $\perp$ to the tangent and so is $||$ to the axes .

It passes through $\bigg(\large\frac{\pi}{6}$$,2\bigg)$.

Therefore the equations of the normal is $x=\large\frac{\pi}{6}$

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