Your derivative is fine.
- this is your gradient at x=1 (in other words the slope of the tangent line at x is y'(x))
The point on is giving the point to use below
You can then use the equation of a straight line with what you found above.
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??)
Your derivative is fine.
- this is your gradient at x=1 (in other words the slope of the tangent line at x is y'(x))
The point on is giving the point to use below
You can then use the equation of a straight line with what you found above.