We first have to differentiate the function to obtain the gradient of the tangent.
Given f(x) = sqrt(x) - e^(sqrt(x))f(x)=√x−e√x.
Then f'(x) = 1/(2sqrt(x))-1/(2sqrt(x))e^sqrt(x).
Now, setting x=1 and substituting into f'(x) we can obtain the gradient:
f'(1) = 1/(2sqrt(1))-1/(2sqrt(1))e^(sqrt(1))=1/2(1-e)
Now we need a y value for the line to pass through. We know as it is tangent to f(x) at x=1 then we can substitute x=1 into f(x) giving us:
f(1) = sqrt(1) - e^sqrt(1) = 1-e
So we now have an x and y value, all that remains is to substitute the values into y= mx+c to find the y intercept.
(1-e) = 1/2(1-e)(1) +c
-> c = 1/2(1-e)
Hence our final answer:
y = 1/2(1-e)x + 1/2(1-e)
If you wish you factor this to tidy it up a bit:
y = 1/2(1-e)(x+1)
Here is a graph showing f(x) (blue) and the normal line at x=1(orange). (The blue line ends at x=0 due to the trouble of taking the square root of a negative).
![enter image source here]()
