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Popular >

d/(dt)(t,t^2,t^3)

Solution

\frac{d}{d t} (t, t^{2}, t^{3})

Solution

(,, (1 \cdot t^{2} t^{3} + 5 t^{4} t))

Solution steps

\frac{d}{d t} (t, t^{2}, t^{3})

Take the constant out:

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

= (,, \frac{d}{d t} (t t^{2} t^{3}))

Apply the Product Rule:

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

f = t, g = t^{2} t^{3}

= (,, (\frac{d t}{d t} t^{2} t^{3} + \frac{d}{d t} (t^{2} t^{3}) t))

\frac{d t}{d t} = 1

\frac{d}{d t} (t^{2} t^{3}) = 5 t^{4}

= (,, (1 \cdot t^{2} t^{3} + 5 t^{4} t))

Popular Examples

limit as x approaches 3-of (2-x)/(x-3)

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

tangent of f(x)=-2x^3,(-2,16)tangent of

f (x) = - 2 x^{3}, (- 2, 16)

(\partial)/(\partial y)(-5/(1+x^2-y))

\frac{\partial}{\partial y} (- \frac{5}{1 + x ^{2} - y})

tangent of f(x)=x^3,\at x=-2 tangent of

f (x) = x^{3}, at x = - 2

integral of 21sec^4(x)

\int 21 sec^{4} (x) d x

Frequently Asked Questions (FAQ)

What is the d/(dt)(t,t^2,t^3) ?
The d/(dt)(t,t^2,t^3) is (,,(1*t^2t^3+5t^4t))
What is the first d/(dt)(t,t^2,t^3) ?
The first d/(dt)(t,t^2,t^3) is (,,(1*t^2t^3+5t^4t))