What is the derivative of (4t^4+4/(t^4))^{1/4} ?

The derivative of (4t^4+4/(t^4))^{1/4} is (4(t^4)^{3/4}(t^8-1))/(t^5(4t^8+4)^{3/4)}

What is the first derivative of (4t^4+4/(t^4))^{1/4} ?

The first derivative of (4t^4+4/(t^4))^{1/4} is (4(t^4)^{3/4}(t^8-1))/(t^5(4t^8+4)^{3/4)}

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derivative of

(4 t^{4} + \frac{4}{t ^{4}})^{\frac{1}{4}}

\frac{4 ( t ^{4} ) ^{\frac{3}{4}} ( t ^{8} - 1 )}{t ^{5} ( 4 t ^{8} + 4 ) ^{\frac{3}{4}}}

Solution steps

\frac{d}{d t} ((4 t^{4} + \frac{4}{t ^{4}})^{\frac{1}{4}})

Simplify

(4 t^{4} + \frac{4}{t ^{4}})^{\frac{1}{4}} : (\frac{4 t ^{8} + 4}{t ^{4}})^{\frac{1}{4}}

= \frac{d}{d t} ((\frac{4 t ^{8} + 4}{t ^{4}})^{\frac{1}{4}})

Apply the chain rule:

\frac{1}{4 ( \frac{4 t ^{8} + 4}{t ^{4}} ) ^{\frac{3}{4}}} \frac{d}{d t} (\frac{4 t ^{8} + 4}{t ^{4}})

= \frac{1}{4 ( \frac{4 t ^{8} + 4}{t ^{4}} ) ^{\frac{3}{4}}} \frac{d}{d t} (\frac{4 t ^{8} + 4}{t ^{4}})

\frac{d}{d t} (\frac{4 t ^{8} + 4}{t ^{4}}) = \frac{16 ( t ^{8} - 1 )}{t ^{5}}

= \frac{1}{4 ( \frac{4 t ^{8} + 4}{t ^{4}} ) ^{\frac{3}{4}}} \cdot \frac{16 ( t ^{8} - 1 )}{t ^{5}}

Simplify

\frac{1}{4 ( \frac{4 t ^{8} + 4}{t ^{4}} ) ^{\frac{3}{4}}} \cdot \frac{16 ( t ^{8} - 1 )}{t ^{5}} : \frac{4 ( t ^{4} ) ^{\frac{3}{4}} ( t ^{8} - 1 )}{t ^{5} ( 4 t ^{8} + 4 ) ^{\frac{3}{4}}}

= \frac{4 ( t ^{4} ) ^{\frac{3}{4}} ( t ^{8} - 1 )}{t ^{5} ( 4 t ^{8} + 4 ) ^{\frac{3}{4}}}

(e^{2 x} + 4) \frac{d y}{d x} = y

f (x) = (cos (x))^{33} x

\frac{d}{d x} (sin (x) cos (y))

\int (x^{2} + x + x^{- 3}) d x

\frac{d}{d x} (\frac{x}{x ^{2} - 9})

What is the derivative of (4t^4+4/(t^4))^{1/4} ?
The derivative of (4t^4+4/(t^4))^{1/4} is (4(t^4)^{3/4}(t^8-1))/(t^5(4t^8+4)^{3/4)}
What is the first derivative of (4t^4+4/(t^4))^{1/4} ?
The first derivative of (4t^4+4/(t^4))^{1/4} is (4(t^4)^{3/4}(t^8-1))/(t^5(4t^8+4)^{3/4)}