What is the derivative of s(t)=48(6t^3-5)^{6/5} ?

The derivative of s(t)=48(6t^3-5)^{6/5} is (5184t^2(6t^3-5)^{1/5})/5

What is the first derivative of s(t)=48(6t^3-5)^{6/5} ?

The first derivative of s(t)=48(6t^3-5)^{6/5} is (5184t^2(6t^3-5)^{1/5})/5

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derivative of

s (t) = 48 (6 t^{3} - 5)^{\frac{6}{5}}

\frac{5184 t ^{2} ( 6 t ^{3} - 5 ) ^{\frac{1}{5}}}{5}

Solution steps

\frac{d}{d t} (48 (6 t^{3} - 5)^{\frac{6}{5}})

Take the constant out:

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

= 48 \frac{d}{d t} ((6 t^{3} - 5)^{\frac{6}{5}})

Apply the chain rule:

\frac{6 ( 6 t ^{3} - 5 ) ^{\frac{1}{5}}}{5} \frac{d}{d t} (6 t^{3} - 5)

= \frac{6 ( 6 t ^{3} - 5 ) ^{\frac{1}{5}}}{5} \frac{d}{d t} (6 t^{3} - 5)

\frac{d}{d t} (6 t^{3} - 5) = 18 t^{2}

= 48 \cdot \frac{6 ( 6 t ^{3} - 5 ) ^{\frac{1}{5}}}{5} \cdot 18 t^{2}

Simplify

48 \cdot \frac{6 ( 6 t ^{3} - 5 ) ^{\frac{1}{5}}}{5} \cdot 18 t^{2} : \frac{5184 t ^{2} ( 6 t ^{3} - 5 ) ^{\frac{1}{5}}}{5}

= \frac{5184 t ^{2} ( 6 t ^{3} - 5 ) ^{\frac{1}{5}}}{5}

e^{x} (5 x^{2} - 10 x + 10)

\frac{d}{d x} (arcsin (sin (17 x)))

\int \frac{x}{x + 60} d x

(3 - x) \frac{d y}{d x} = y

What is the derivative of s(t)=48(6t^3-5)^{6/5} ?
The derivative of s(t)=48(6t^3-5)^{6/5} is (5184t^2(6t^3-5)^{1/5})/5
What is the first derivative of s(t)=48(6t^3-5)^{6/5} ?
The first derivative of s(t)=48(6t^3-5)^{6/5} is (5184t^2(6t^3-5)^{1/5})/5