What is (d^2)/(dx^2)((y(t,x))t^{-1}) ?

The answer to (d^2)/(dx^2)((y(t,x))t^{-1}) is t^{-1} derivative of d/(dx({y}(t,x)))

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(d^2)/(dx^2)((y(t,x))t^{-1})

\frac{d ^{2}}{d x ^{2}} ((y (t, x)) t^{- 1})

t^{- 1} \frac{d}{d x} (\frac{d}{d x} (y (t, x)))

Solution steps

\frac{d ^{2}}{d x ^{2}} ((y (t, x)) t^{- 1})

\frac{d}{d x} (y (t, x) t^{- 1}) = t^{- 1} \frac{d}{d x} (y (t, x))

= \frac{d}{d x} (t^{- 1} \frac{d}{d x} (y (t, x)))

\frac{d}{d x} (t^{- 1} \frac{d}{d x} (y (t, x))) = t^{- 1} \frac{d}{d x} (\frac{d}{d x} (y (t, x)))

= t^{- 1} \frac{d}{d x} (\frac{d}{d x} (y (t, x)))

2 \cdot 2 x^{\frac{1}{2}}

- \frac{9}{5} e^{- 25 t}

3 (3 - 3 y)^{2} y^{2}

e^{- \frac{x}{3}} d x

1.67 \cdot 1 0^{- 6}

What is (d^2)/(dx^2)((y(t,x))t^{-1}) ?
The answer to (d^2)/(dx^2)((y(t,x))t^{-1}) is t^{-1} derivative of d/(dx({y}(t,x)))