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Find an equation of the tangent line to the curve $y = sin(3x) + sin^2 (3x)$ given the point (0,0). Answer is $y = 3x$, but please explain solution steps.

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Hint:

Do you know that the slope of the tangent line at a point of the graph of a function is the derivative of the function at this point?

So, for $y = \sin(3x) + \sin^2 (3x)$ find the derivative $y'$ ( can you do?), then evaluate this derivative for $x=0$

Now the line has equation $y=mx$ with $m=y'(0)$


Using the chain rule the derivative is: $$ y'=\cos(3x)\cdot(3x)'+2\sin(3x)(\sin(3x))'=3\cos(3x)+2\sin(3x)\cos(3x)(3x)'$$$$=3\cos(3x)+6\sin(3x)\cos(3x) $$ so $y'(0)=3$.

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  • $\begingroup$ I know that I have to find the derivative, but I don't know how to. I also know that once you find m (by taking the derivative) you can plug it into the $y-y=m(x-x)$ or $y = mx + b$ slope formulas. $\endgroup$ – Chaniqua Ranson Oct 5 '16 at 2:04
  • $\begingroup$ Added to my answer. $\endgroup$ – Emilio Novati Oct 5 '16 at 7:40
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Goal

Find the slope $m$, and intercept $b$, for the line $$ y = mx + b, $$ tangent at the origin to the curve $$ f(x) = \sin ^2(3 x)+\sin (3 x). $$

Intercept $b$

Because the function goes through the origin, the tangent line will also go through the origin. Therefore the $y-$intercept $b=0$.

Slope $m$

The slope of the tangent line $m$ is, by definition, the same as the slope of the target function at the point of contact. The slope of the function is $$ f'(x) = 6 \sin (3 x) \cos (3 x) + 3 \cos (3 x). $$ The problem specifies $x_{*} = 0.$ The point of contact is $$ \left( x_{*}, f(x_{*}) \right) = \left( 0, 3 \right). $$ The slope of the function at the point of contact is $$ f'( x_{*} ) = 6 \sin (3 x_{*}) \cos (3 x_{*}) + 3 \cos (3 x_{*}) = 6\cdot 0 \cdot 0 + 3 \cdot 1 = 3. $$ The slope of the tangent line is $$ m = f'( x_{*} ) = 3. $$

Solution The equation of the line tangent to $f(x)$ at $x_{*} = 0$ is $$ y = mx + b = 3x. $$

Tangent line

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