Math Help - equation of tangent

1. equation of tangent

Given that $e^{tan^{-1}x}=1+x+\frac{1}{2}x^2+...$ and $(1+3x)^{-\frac{1}{2}}=1-\frac{3}{2}x+\frac{27}{8}x^2+...,$deduce the equation of the tangent to the curve $y=\frac{e^{tan^{-1}x}}{\sqrt{1+3x}}$

2. Re: equation of tangent

Originally Posted by Punch
$e^{tan^{-1}x}=1+x+\frac{1}{2}x^2+...$
Actually, $e^x=1+x+\frac{1}{2}x^2+...$

deduce the equation of the tangent to the curve $y=\frac{e^{tan^{-1}x}}{\sqrt{1+3x}}$
Write it as $y=(1+3x)^{-\frac{1}{2}}e^{tan^{-1}x}$ and use the product rule. (Make sure you use the correct formula for $e^{tan^{-1}x}$.)

3. Re: equation of tangent

Originally Posted by alexmahone
Actually, $e^x=1+x+\frac{1}{2}x^2+...$

Write it as $y=(1+3x)^{-\frac{1}{2}}e^{tan^{-1}x}$ and use the product rule. (Make sure you use the correct formula for $e^{tan^{-1}x}$.)
So i will be able to find the expression of dy/dx. Next, ill need to use y-y1=m(x-x1) but what coordinates do i use for y1 and x1?

4. Re: equation of tangent

Originally Posted by Punch
So i will be able to find the expression of dy/dx. Next, ill need to use y-y1=m(x-x1) but what coordinates do i use for y1 and x1?
y(0) = 1. So you may take y1 = 1 and x1 = 0.

5. Re: equation of tangent

Originally Posted by alexmahone
y(0) = 1. So you may take y1 = 1 and x1 = 0.
But the equation of the tangent at different point are different. But question did not state which point of the curvethe tangent is required. Can i assume it is asking for the point u stated?

6. Re: equation of tangent

Originally Posted by Punch
But question did not state which point of the curvethe tangent is required.
Unless the question states this, it cannot be answered.

Can i assume it is asking for the point u stated?
No.