Question

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

Answer 1

The slope of the tangent is `e`.

Explanation:

For functions of the form `y= e^(f(x))`, the derivative is given by `y' = f'(x)e^(f(x))`. Also, the derivative of `lnx = 1/x`. We cannot forget that we are differentiating with respect to `x`.

`cosx(e^sinx) = 1/y(dy/dx)`

`(cosx(e^sinx))/(1/y) = dy/dx`

`ycosxe^(sinx) = dy/dx`

We could alternatively substitute `e^(e^(sinx))` for `y`:

`e^(e^sinx)cosxe^(sinx) = dy/dx`

This can be combined.

`e^(e^sinx + sinx)cosx = dy/dx`

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

`e^(e^sin(0) + sin(0))cos(0) = m_"tangent"`

`e^(e^0 + 0)(1) = m_"tangent"`

`e^(1 + 0)(1) = m_"tangent"`

`e = m_"tangent"`

Hence, the slope of the tangent is `e`.

Hopefully this helps!