How can I go about finding a tangent line for the following:

(x^3)*(R^5)=1 at the point (x,R)=(1,1)

I took the derivative of the equation but I didn't know where to go after that.

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- Oct 21st 2012, 12:13 PMsmgrojeanFinding a Tangent Line
How can I go about finding a tangent line for the following:

(x^3)*(R^5)=1 at the point (x,R)=(1,1)

I took the derivative of the equation but I didn't know where to go after that. - Oct 21st 2012, 12:40 PMskeeterRe: Finding a Tangent Line
$\displaystyle \frac{dR}{dx}$ is the slope of the curve at any point $\displaystyle (x,R)$ on the curve ...

evaluate the derivative at $\displaystyle x = 1$ , $\displaystyle R = 1$ to find the slope of the tangent line at that specific point, then use the point-slope form of a linear equation ...

$\displaystyle R - R_1 = m(x - x_1)$ - Oct 23rd 2012, 03:28 PMsmgrojeanRe: Finding a Tangent Line
So if I take the derivative I get:

3*R^5*X^2

Then if I evaluate that at x=1, R=1 I get:

3*1^5*1^2 = 3

So the point slope of the linear equation should be the following:

R - 1 = 3 ( X - 1) - Oct 23rd 2012, 05:10 PMProve ItRe: Finding a Tangent Line
That is not the derivative. Since R is a function of x, you will need to use implicit differentiation.

- Oct 23rd 2012, 05:25 PMskeeterRe: Finding a Tangent Line
$\displaystyle x^3 \cdot R^5 = 1$

solve for $\displaystyle R$ in terms of $\displaystyle x$ ...

$\displaystyle R = x^{-\frac{3}{5}}$

... now find $\displaystyle \frac{dR}{dx}$ - Oct 26th 2012, 12:37 PMsmgrojeanRe: Finding a Tangent Line
so the derivative would actually be:

dr/dx = -3/(5*x^(8/5)) - Oct 26th 2012, 01:53 PMskeeterRe: Finding a Tangent Line
ok ... find the equation of the tangent line.

- Oct 28th 2012, 11:26 AMsmgrojeanRe: Finding a Tangent Line
so the equation should be:

R - 1 = -3/5(x-1)