I am given the function f(x) = $\sqrt{3x+5}$
I have calculated the expression of the nth derivative to be
$f^{(n)}(x)=3^n\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2n-3}{2}\right)(3x+5)^{-(2n-1)/2}$
How would I prove this expression to be true by induction?
The base case is $n = 1$, which is true by
$f'(x) = \frac 32 (3x+5)^{-1/2} = 3^1\frac 12(3x+5)^{-(2-1)/2}$
Now differentiate $f^{(n)}(x)$, and if the resulting expression is $f^{(n+1)}(x)=3^{(n+1)}\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2(n+1)-3}{2}\right)(3x+5)^{-(2(n+1)-1)/2}$
Then you've proven that the expression is true, by induction.