What is the derivative of 41(6t^3-7)^{6/5} ?

The derivative of 41(6t^3-7)^{6/5} is (4428t^2(6t^3-7)^{1/5})/5

What is the first derivative of 41(6t^3-7)^{6/5} ?

The first derivative of 41(6t^3-7)^{6/5} is (4428t^2(6t^3-7)^{1/5})/5

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derivative of

41 (6 t^{3} - 7)^{\frac{6}{5}}

\frac{4428 t ^{2} ( 6 t ^{3} - 7 ) ^{\frac{1}{5}}}{5}

Solution steps

\frac{d}{d t} (41 (6 t^{3} - 7)^{\frac{6}{5}})

Take the constant out:

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

= 41 \frac{d}{d t} ((6 t^{3} - 7)^{\frac{6}{5}})

Apply the chain rule:

\frac{6 ( 6 t ^{3} - 7 ) ^{\frac{1}{5}}}{5} \frac{d}{d t} (6 t^{3} - 7)

= \frac{6 ( 6 t ^{3} - 7 ) ^{\frac{1}{5}}}{5} \frac{d}{d t} (6 t^{3} - 7)

\frac{d}{d t} (6 t^{3} - 7) = 18 t^{2}

= 41 \cdot \frac{6 ( 6 t ^{3} - 7 ) ^{\frac{1}{5}}}{5} \cdot 18 t^{2}

Simplify

41 \cdot \frac{6 ( 6 t ^{3} - 7 ) ^{\frac{1}{5}}}{5} \cdot 18 t^{2} : \frac{4428 t ^{2} ( 6 t ^{3} - 7 ) ^{\frac{1}{5}}}{5}

= \frac{4428 t ^{2} ( 6 t ^{3} - 7 ) ^{\frac{1}{5}}}{5}

\int_{0}^{\frac{π}{2}} t sin (2 t) d t

y = arctan (2 x)

cos (θ), sin (2 θ)

\frac{\partial}{\partial x} (sin (5 x + 2 y))

\int e^{t} + 2 t + ln (t) d t

What is the derivative of 41(6t^3-7)^{6/5} ?
The derivative of 41(6t^3-7)^{6/5} is (4428t^2(6t^3-7)^{1/5})/5
What is the first derivative of 41(6t^3-7)^{6/5} ?
The first derivative of 41(6t^3-7)^{6/5} is (4428t^2(6t^3-7)^{1/5})/5