What is the d/(dt)(5t^2x) ?

The d/(dt)(5t^2x) is 5(2tx+(dx)/(dt)t^2)

What is the first d/(dt)(5t^2x) ?

The first d/(dt)(5t^2x) is 5(2tx+(dx)/(dt)t^2)

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\frac{d}{d t} (5 t^{2} x)

5 (2 t x + \frac{d x}{d t} t^{2})

Solution steps

\frac{d}{d t} (5 t^{2} x)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d}{d t} (t^{2} x)

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t^{2}, g = x

= 5 (\frac{d}{d t} (t^{2}) x + \frac{d x}{d t} t^{2})

\frac{d}{d t} (t^{2}) = 2 t

= 5 (2 t x + \frac{d x}{d t} t^{2})

\int u^{4} d u

x \to 0 - lim (\frac{x ^{2}}{3} - \frac{1}{x})

y^{^{'''}} + 6 y^{^{''}} + 45 y^{^{'}} = 0

y - x + x y^{^{'}} = 0

\int_{3}^{x^{3}} t^{4} d t

What is the d/(dt)(5t^2x) ?
The d/(dt)(5t^2x) is 5(2tx+(dx)/(dt)t^2)
What is the first d/(dt)(5t^2x) ?
The first d/(dt)(5t^2x) is 5(2tx+(dx)/(dt)t^2)