What is the derivative of m(t)=5t(3t^4-1)^5 ?

The derivative of m(t)=5t(3t^4-1)^5 is 5((3t^4-1)^5+60t^4(3t^4-1)^4)

What is the first derivative of m(t)=5t(3t^4-1)^5 ?

The first derivative of m(t)=5t(3t^4-1)^5 is 5((3t^4-1)^5+60t^4(3t^4-1)^4)

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derivative of

m (t) = 5 t (3 t^{4} - 1)^{5}

5 ((3 t^{4} - 1)^{5} + 60 t^{4} (3 t^{4} - 1)^{4})

Solution steps

\frac{d}{d t} (5 t (3 t^{4} - 1)^{5})

Take the constant out:

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

= 5 \frac{d}{d t} (t (3 t^{4} - 1)^{5})

Apply the Product Rule:

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

f = t, g = (3 t^{4} - 1)^{5}

= 5 (\frac{d t}{d t} (3 t^{4} - 1)^{5} + \frac{d}{d t} ((3 t^{4} - 1)^{5}) t)

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

\frac{d}{d t} ((3 t^{4} - 1)^{5}) = 60 t^{3} (3 t^{4} - 1)^{4}

= 5 (1 \cdot (3 t^{4} - 1)^{5} + 60 t^{3} (3 t^{4} - 1)^{4} t)

Simplify

5 (1 \cdot (3 t^{4} - 1)^{5} + 60 t^{3} (3 t^{4} - 1)^{4} t) : 5 ((3 t^{4} - 1)^{5} + 60 t^{4} (3 t^{4} - 1)^{4})

= 5 ((3 t^{4} - 1)^{5} + 60 t^{4} (3 t^{4} - 1)^{4})

\frac{d}{d x} (x cos (y) - y e^{x})

f (x) = 6 x^{5} e^{x} + e^{x} x^{6}

\int tan (x) [ln (cos (x))] d x

f (x) = cos (θ^{2})

x \to 5 lim (\frac{x ^{2} - 7 x + 10}{x ^{2} - 25})

What is the derivative of m(t)=5t(3t^4-1)^5 ?
The derivative of m(t)=5t(3t^4-1)^5 is 5((3t^4-1)^5+60t^4(3t^4-1)^4)
What is the first derivative of m(t)=5t(3t^4-1)^5 ?
The first derivative of m(t)=5t(3t^4-1)^5 is 5((3t^4-1)^5+60t^4(3t^4-1)^4)