`y = e^sinhx , (0, 1)` Find an equation of the tangent line to the graph of the function at the given point

Expert Answers

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The tangent line must go through the given point `(x_0, y_0)` and have the slope of `y'(x_0).` Thus the equation of the tangent line is

`(y - y_0) = (x - x_0)*y'(x_0).`

To find the derivative we need the chain rule and the derivative of `sinh(x),` which is `cosh(x).` Therefore `y'(x) = (e^sinh(x))' =e^sinh(x)*cosh(x).` For `x = x_0 = 0` it is `e^0*1 = 1.`

So the equation of the tangent line is  `y - 1 = (x - 0)*1,` or simply y = x + 1.

We have to check that `y(x_0) = x_0.`  Yes, `y(x_0) = y(0) = 1` and `x_0 = 1.`

 

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