[SOLVED] equation of a tangent line

**simply07** · November 4th 2007, 04:27 PM

Having a little bit of problem....
Let c be a nonzero real number. What is the equation of the tangent line to the graphof the function y=sin(x/x+c) at the point x=0.

i get the derivate of cos(1/1+c) but i really dont think that it right!!! am i am differentiating properly!!!???

**Plato** · November 4th 2007, 04:45 PM

$\frac{d}{{dx}}\left( {\sin \left( {\frac{x}{{x + c}}} \right)} \right) = \cos \left( {\frac{x}{{x + c}}} \right)\left( {\frac{c}{{\left( {x + c} \right)^2 }}} \right)$

**Soroban** · November 4th 2007, 04:46 PM

simply07!

Let $c$ be a nonzero real number.
What is the equation of the tangent line to the graph of the function
$y \:=\:\sin\left(\frac{x}{x+c}\right)$ at the point $x=0.$

i get the derivate of $\cos\left(\frac{1}{1+c}\right)$ but i really dont think that it right!
am i differentiating properly? . . . . no

You'd better review the Chain Rule ... and the Quotient Rule.

**simply07** · November 4th 2007, 05:22 PM

ok i see the mistake... so now to find the value at x = 0, i would simply replace x for 0....

then my answer woul be 1/c ???
this make sence but is it right?

thanks

**topsquark** · November 4th 2007, 06:10 PM

Originally Posted by simply07

ok i see the mistake... so now to find the value at x = 0, i would simply replace x for 0....

then my answer woul be 1/c ???
this make sence but is it right?

thanks

That's the correct slope of the tangent line. So your tangent line has the slope-intercept form of:
$y = \frac{1}{c} \cdot x + b$

It must pass through the point (0, y(0)), so plug this point in to find b.

-Dan

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