Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find an equation for the tangent line to the curve

$$x\sin(xy-y^2)=x^2-1$$

through the point $(1,1)$.

share|improve this question
2  
Do you have any thoughts on the problem? –  T. Bongers Sep 19 '13 at 5:22

3 Answers 3

Hint: You need to find the equation of the line

$$ \frac{y-y_0}{x-x_0}=m, $$

where $m$ is the slope at the point $(x_0,y_0)$ which is given by

$$ m = \frac{dy}{dx}\Big|_{(x,y)=(x_0,y_0)}. $$

share|improve this answer

First of all make sure that your equation defines $y$ with respect to $x$. This allows you to know that the equation defines $y$ with respect to $x$ implicitly and so you can do the implicit differentiation. After that use the following way to find the certain slope $y'|_{x=1}$. If $F(x,y)=0$ defines $y$ with respect to $x$ implicitly, then $$y'=\frac{-F_x}{F_y}$$

share|improve this answer
    
This needs a TU! +1 –  amWhy Sep 19 '13 at 12:32

When the curve is given by $y = f(x)$ then the slope of the tangent is $\frac{dy}{dx}$,then the equation of the tangent line can be found by the equation.$$y=\frac{dy}{dx}x+c$$the value of $c$ can be found with the help of the given point $(1,1)$. given:$$x\sin(xy-y^2)=x^2-1$$$$y^2-xy+arcsin\left(\frac{x^2-1}{x}\right)=0$$solve this quadratic equation in $y$ to get$$y=\frac{x+\sqrt{x^2-4arcsin\left(\frac{x^2-1}{x}\right)}}{2}$$the other root cancels as it does not satisfy $x=1,y=1$.find $\frac{dy}{dx}_{(1,1)}$$$\frac{dy}{dx}=\frac{1}{2}+\frac{2x-4\left(1+\frac{1}{x^2}\right)\left(\frac{1}{\sqrt{1-\frac{x^2-1}{x}}}\right)}{4\sqrt{x^2-4arcsin\left(\frac{x^2-1}{x}\right)}}$$$$\frac{dy}{dx}_{(1,1)}=-1$$therefore the equation of the tangent is$$y=2-x$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.