Hans Georg Schaathun
August 2016
$$h(t) = 10 - 4{,}9t^2$$
$$h^{-1}(y) = \sqrt{\frac{10-y}{4{,}9}}$$
$$h(t) = 10 - 4{,}9t^2 = y$$
$$\frac{d}{dy}\big[10 - 4{,}9t^2\big] = \frac{d}{dy}y$$
$$\frac{d}{dy}\big[10 - 4{,}9t^2\big] = 1$$
$$- 4{,}9\frac{d}{dt} t^2 \frac{dt}{dy} = 1$$
$$- 9{,}8t\frac{dt}{dy} = 1$$
$$\frac{dt}{dy} = - \frac1{9{,}8t}$$
$$h(t) = y$$
$$\frac{d}{dy}h(t) = \frac{d}{dy}y$$
$$\frac{d}{dy}h(t) = 1$$
$$h'(t)\frac{dt}{dy} = 1$$
$$h'(t)\frac{d}{dy}h^{-1}(y) = 1$$
$$\frac{d}{dy}h^{-1}(y) = \frac1{h'(h^{-1}(y))}$$