What is the d/(dt)(1+(t-y)^2) ?

The d/(dt)(1+(t-y)^2) is 2(t-y)(-(dy)/(dt)+1)

What is the first d/(dt)(1+(t-y)^2) ?

The first d/(dt)(1+(t-y)^2) is 2(t-y)(-(dy)/(dt)+1)

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

2 (t - y) (- \frac{d y}{d t} + 1)

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d t} (1) = 0

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

= 0 + 2 (t - y) (1 - \frac{d y}{d t})

Simplify

= 2 (t - y) (- \frac{d y}{d t} + 1)

What is the d/(dt)(1+(t-y)^2) ?
The d/(dt)(1+(t-y)^2) is 2(t-y)(-(dy)/(dt)+1)
What is the first d/(dt)(1+(t-y)^2) ?
The first d/(dt)(1+(t-y)^2) is 2(t-y)(-(dy)/(dt)+1)