# How do you find the slope of the line tangent to #e^sinx=lny# at #(0,e)#?

The slope of the tangent is

This can be combined.

The slope of the tangent can be found by evaluating your x-coordinate within the derivative.

Hopefully this helps!

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To find the slope of the line tangent to the curve represented by the equation e^sinx = lny at the point (0,e), we can use the concept of implicit differentiation.

First, we differentiate both sides of the equation with respect to x.

For the left side, we use the chain rule: d/dx(e^sinx) = e^sinx * cosx.

For the right side, we use the chain rule and the fact that d/dx(lny) = 1/y: d/dx(lny) = 1/y * dy/dx.

Now, we can substitute the values of x = 0 and y = e into the derived equation.

At x = 0, sinx = 0, and cosx = 1.

At y = e, we have: e^sinx = lny = e^0 = 1.

Substituting these values, we get: 1 * 1 = 1/y * dy/dx.

Simplifying, we find: dy/dx = 1/y.

At the point (0,e), y = e.

Therefore, the slope of the line tangent to the curve at (0,e) is equal to 1/e.

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