Hence the slope of the graph of the square function at the point (3, 9) is 6, and so its derivative at x = 3 is f ′ (3) = 6 More generally, a similar computation shows that the derivative of the square function at x = a is f ′ (a) = 2aF (x)=x^3 f (x)=\ln (x5) f (x)=\frac {1} {x^2} y=\frac {x} {x^26x8} f (x)=\sqrt {x3} f (x)=\cos (2x5) f (x)=\sin (3x) functionscalculator f\left (x\right)=x^3For example, if $f(x) = x^3 x 1$, $f^{1}(3) = 1$ because $f(1) = 3$ $\endgroup$ – Reese May 22 '18 at 1724 $\begingroup$ You would have to solve
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