Tangent line to a curve.

**wintermath** · June 19th 2011, 06:46 AM

I'm trying to find the equation of the line tangent to the curve of y=ln(sqrt(7x^2+9)) at x=1.

So I need to find the first derivative:

y'= 7x/(7x^2+9) and at x=1 I find y= 16 (1,16)

Because y= (approx.) 2.77 at x=1 I have a point to substitute in my equation of the line:

y=16x+b

and I find b= -16.23 (approx.)

then the equation of my tangent line at 1 is y=16x-13.23

except that looks nothing like the tangent line on my graph. Where am I off (way off??)

**e^(i*pi)** · June 19th 2011, 06:53 AM

Your derivative is fine.

$y'(1) = \dfrac{7}{16}$ - this is your gradient at x=1 (in other words the slope of the tangent line at x is y'(x))

The point on $f(x)$ is $f(1) = \ln(\sqrt{7+9}) = \ln(4) = 2\ln(2)$ giving the point $(1, 2\ln(2))$ to use below

You can then use the equation of a straight line with what you found above.

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