Saturday

August 30, 2014

August 30, 2014

Posted by **Aurora** on Monday, June 11, 2007 at 8:26pm.

Find the equation of the tangent line to the curve 4e^xy = 2x + y at point (0,4).

On the left side, is the "xy" the exponent of e, or do you mean 4(e^x)y ?

Use implicit differentiation to get the slope m = dy/dx at x=0, y=4.

If 4 e^(xy) = 2x + y, then

4 e^(xy)* (y + x dy/dx) = 2 + dy/dx

Solve for dy/dx = m. The tangent line equation is then

(y-4) = m x

the exponent of e is "xy".

It's must clearer now, thank you.

In that case, my derivation should be correct. Substitute x=0, y=4 in the equation

4 e^(xy)* (y + x dy/dx) = 2 + dy/dx

4*1*(4 + 0) = 2 + dy/dx

dy/dx = 14

y-4 = 14 x

y = 14 x + 4

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