Find the equation of the tangent line to the graph

**SammyS** · January 16th 2011, 07:46 PM

Originally Posted by pickslides

You can leave it as $\displaystyle x=x_0$ for a future value, then $\displaystyle y-f(x_0)=f'(x_0)(x-x_0)$

Hi zhangxupage2.

Do as pickslides suggests. Substitute the following.

$f(x_0)=3x_0 \tan^2(x_0) + (x_0)^2 \cos(x_0)$

and as you found, $f'(x_0)=3\tan^2(x_0)+6\tan(x_0)\sec^2(x_0) + 2x_0\cos(x_0)-(x_0)^2\sin(x_0)$

**zhangxupage2** · January 16th 2011, 07:50 PM

Thanks a lot everyone!

**zhangxupage2** · January 16th 2011, 08:24 PM

What if the question asks me to pick a random value and plug it in, and the find the equation. What is an easy number to pick, considering that I can't use a calculator?

**dwsmith** · January 16th 2011, 08:27 PM

Originally Posted by zhangxupage2

What if the question asks me to pick a random value and plug it in, and the find the equation. What is an easy number to pick, considering that I can't use a calculator?

Values associated with the unit circle.

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