I need to find the square root of: $\displaystyle f(x) = \sqrt{x^2+15} $ The answer I got is: $\displaystyle f'(x) = \frac{x}{\sqrt{x^2+15}} $ Is this correct? if not please explain.
Follow Math Help Forum on Facebook and Google+
the answer is actually: f'(x) = x / 2*(sqrt(x^2+15) because: f(x) = (x^2+15)^(1/2) so: f'(x) = (1/2)*(x^2+15)^(-1/2)*x <------ the inner derivative and finally: f'(x) = x / 2*(sqrt(x^2+15)
wouldn't the inner derivative be 2x? since it was x^2
my bad... your answer was correct...
great...thank you
View Tag Cloud