Find an equation of the tangent line at the given point. $7y^2 − xy^2 − x^3 =0$ the point is $(\frac72,\frac72)$
Ive found the derivative: $14y\frac{dy}{dx}-y^2-2yx\frac{dy}{dx}-3x^2=0$
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I found the derivative and then plugged in the points to find the slope. I don't know if that is right. – Kels Jul 01 '14 at 00:07
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Can you write this out in your question? It's much easier for people to help you if they know exactly where you need help! – Mathmo123 Jul 01 '14 at 00:07
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Yes I will. Thanks! – Kels Jul 01 '14 at 00:09
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The derivative looks right to me! Do you know what to do next? – Mathmo123 Jul 01 '14 at 00:11
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No idea.. Thats my problem – Kels Jul 01 '14 at 00:12
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So you have a point on the line and the slope. Shouldn't you be able to get an equation out of that? – Mike Jul 01 '14 at 00:13
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Kelsey: I recommend you to use \cfrac{dy}{dx} or \dfrac{dy}{dx} instead of \frac{dy}{dx}. $$\cfrac{dy}{dx} $$ and $$\frac{dy}{dx}$$ $$\text{Notice the difference.}$$ – Kushashwa Ravi Shrimali Jul 01 '14 at 00:14
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That was my correction! Sorry. But thanks for that! – Mathmo123 Jul 01 '14 at 00:14
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okay thanks and I guess but what about the dy/dx – Kels Jul 01 '14 at 00:14
1 Answers
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I'll start you off: the gradient of the tangent line is given by the value of $\frac{dy}{dx}$ at that point.
1) Sub in the values for $x$ and $y$ and solve for $\frac{dy}{dx}$ to get the gradient.
2) Once you have the gradient $m$, you need to find the $y$-intercept. Let me know if you need a hint for that.
Mathmo123
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Thanks haha! I was plugging in the wrong number and I was just getting super frustrated and then you helped.. – Kels Jul 01 '14 at 00:21