# What is the equation of the line normal to # f(x)=sqrt(x)-e^(sqrtx)# at # x=1#?

We first have to differentiate the function to obtain the gradient of the tangent.

Given

Then

Now, setting

Now we need a

So we now have an

Hence our final answer:

If you wish you factor this to tidy it up a bit:

Here is a graph showing

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The equation of the line normal to f(x) = sqrt(x) - e^(sqrt(x)) at x = 1 is y = -2x + 2.

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