What is the derivative of 3t(2t^2-5)^4 ?

The derivative of 3t(2t^2-5)^4 is 3((2t^2-5)^4+16t^2(2t^2-5)^3)

What is the first derivative of 3t(2t^2-5)^4 ?

The first derivative of 3t(2t^2-5)^4 is 3((2t^2-5)^4+16t^2(2t^2-5)^3)

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derivative of

3 t (2 t^{2} - 5)^{4}

3 ((2 t^{2} - 5)^{4} + 16 t^{2} (2 t^{2} - 5)^{3})

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = t, g = (2 t^{2} - 5)^{4}

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

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

\frac{d}{d t} ((2 t^{2} - 5)^{4}) = 16 t (2 t^{2} - 5)^{3}

= 3 (1 \cdot (2 t^{2} - 5)^{4} + 16 t (2 t^{2} - 5)^{3} t)

Simplify

3 (1 \cdot (2 t^{2} - 5)^{4} + 16 t (2 t^{2} - 5)^{3} t) : 3 ((2 t^{2} - 5)^{4} + 16 t^{2} (2 t^{2} - 5)^{3})

= 3 ((2 t^{2} - 5)^{4} + 16 t^{2} (2 t^{2} - 5)^{3})

\frac{( s - 2 )}{( s - 2 ) ^{2} + 1}

x^{9} d x + y^{9} d y = 0

\frac{1}{s ( s ^{2} + 2 s + 1 )}

\frac{\partial}{\partial x} (e^{2 x + y})

y = f (x) = tan (x + 1)

What is the derivative of 3t(2t^2-5)^4 ?
The derivative of 3t(2t^2-5)^4 is 3((2t^2-5)^4+16t^2(2t^2-5)^3)
What is the first derivative of 3t(2t^2-5)^4 ?
The first derivative of 3t(2t^2-5)^4 is 3((2t^2-5)^4+16t^2(2t^2-5)^3)