What is (d^2)/(dx^2)((y(t,x))(1+t^2)) ?

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

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(d^2)/(dx^2)((y(t,x))(1+t^2))

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

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

Solution steps

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

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

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

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

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

(5 - 6 x)^{2}

\frac{( 5 - x ) ^{7} ( x - 13 ) ^{2}}{( x ^{2} + 7 ) ^{2}}

x^{2} + (- y - 2)^{2} 0.2

5 (e^{2 x} + x)^{4}

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