If $r(x)=\sin^2(3x)$, find the slope of the tangent line at $x=\frac{\pi}{6}$

Key concept: The slope of the tangent line is the derivative of the function at the given x-value. Here, the chain rule is the key step.

Step by step

We need the slope of the tangent line to

$r(x)=\sin^2(3x)$

at $x=\frac{\pi}{6}$.

Step 1: Differentiate the function

Rewrite as

$r(x)=[\sin(3x)]^2$

Use the chain rule:

$r'(x)=2\sin(3x)\cdot \cos(3x)\cdot 3$

$r'(x)=6\sin(3x)\cos(3x)$

You can also write this as

$r'(x)=3\sin(6x)$

using the double-angle identity.

Step 2: Evaluate at $x=\frac{\pi}{6}$

$r'\left(\frac{\pi}{6}\right)=3\sin\left(6\cdot \frac{\pi}{6}\right)=3\sin(\pi)=0$

Answer

$\boxed{0}$

So the slope of the tangent line is $0$.

Pitfall alert

Be careful not to plug $x=\frac{\pi}{6}$ into the original function and call that the slope. The slope comes from the derivative, not the function value. Also, using $\sin^2(3x)$ means the square applies to the sine value, not to $3x$.

Try different conditions

If the point were given as a y-value instead of an x-value, you would first identify the matching x-coordinate on the curve and then evaluate the derivative there. If the function were $\cos^2(3x)$, the derivative method would be the same, but the trig derivative would change sign.

Further reading

tangent line, chain rule, derivative