What is the derivative of s(t)=\sqrt[4]{2t^2-4t+7} ?

The derivative of s(t)=\sqrt[4]{2t^2-4t+7} is (t-1)/((2t^2-4t+7)^{3/4)}

What is the first derivative of s(t)=\sqrt[4]{2t^2-4t+7} ?

The first derivative of s(t)=\sqrt[4]{2t^2-4t+7} is (t-1)/((2t^2-4t+7)^{3/4)}

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derivative of

\frac{t - 1}{( 2 t ^{2} - 4 t + 7 ) ^{\frac{3}{4}}}

Solution steps

Apply the chain rule:

\frac{1}{4 ( 2 t ^{2} - 4 t + 7 ) ^{\frac{3}{4}}} \frac{d}{d t} (2 t^{2} - 4 t + 7)

= \frac{1}{4 ( 2 t ^{2} - 4 t + 7 ) ^{\frac{3}{4}}} \frac{d}{d t} (2 t^{2} - 4 t + 7)

\frac{d}{d t} (2 t^{2} - 4 t + 7) = 4 t - 4

= \frac{1}{4 ( 2 t ^{2} - 4 t + 7 ) ^{\frac{3}{4}}} (4 t - 4)

Simplify

\frac{1}{4 ( 2 t ^{2} - 4 t + 7 ) ^{\frac{3}{4}}} (4 t - 4) : \frac{t - 1}{( 2 t ^{2} - 4 t + 7 ) ^{\frac{3}{4}}}

= \frac{t - 1}{( 2 t ^{2} - 4 t + 7 ) ^{\frac{3}{4}}}

What is the derivative of s(t)=\sqrt[4]{2t^2-4t+7} ?
The derivative of s(t)=\sqrt[4]{2t^2-4t+7} is (t-1)/((2t^2-4t+7)^{3/4)}
What is the first derivative of s(t)=\sqrt[4]{2t^2-4t+7} ?
The first derivative of s(t)=\sqrt[4]{2t^2-4t+7} is (t-1)/((2t^2-4t+7)^{3/4)}