# Find the equation of the line tangent to the graph of f at the indicated value of x. (√=square root) f(x)=√lnx ; x=e f(x)=e√x ; x=1 Thank you!

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`f(x) = sqrt(ln(x))` , x = e

Using the chain rule,

`f'(x) = 1/x*1/(2sqrt(ln(x))) `

`f'(e) = 1/e*1/(2sqrt(ln(e))) = 1/e*1/(2sqrt(1)) = 1/e*1/2 = 1/(2e)`

`f(e) = sqrt(ln(e)) = sqrt(1) = 1`

So the tangent is a line with slope `1/(2e)` at point (e,1)

Using point slope we find

`y - 1 = 1/(2e)(x - e)`

`y = 1/(2e)x-1/2+1`

`y = 1/(2e)x+1/2`

The second equation

`f(x) = e^sqrt(x)` at `x = 1`

`f(1) = e^sqrt(1) = e^1 = e`

So our point is (1,e)

f(x) we can differentiate using the chain rule to get

`f'(x)=e^sqrt(x) * 1/(2sqrt(x))`

`f'(1) = e^sqrt(1) * 1/(2sqrt(1)) = e^1 * 1/2 = 1/2e`

Now we can use point slope to get

`y - e = 1/2e(x - 1)`

`y = 1/2ex - 1/2e + e = 1/2ex + 1/2e`

So the answers are (1) `y = 1/(2e) x + 1/2` and (2) `y = 1/2ex+1/2e`

thank you very much!