What is the derivative of \sqrt[4]{(2x+x^5)^7} ?

The derivative of \sqrt[4]{(2x+x^5)^7} is (7(2x+x^5)^6(2+5x^4))/(4((2x+x^5)^7)^{3/4)}

What is the first derivative of \sqrt[4]{(2x+x^5)^7} ?

The first derivative of \sqrt[4]{(2x+x^5)^7} is (7(2x+x^5)^6(2+5x^4))/(4((2x+x^5)^7)^{3/4)}

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derivative of

\frac{7 ( 2 x + x ^{5} ) ^{6} ( 2 + 5 x ^{4} )}{4 ( ( 2 x + x ^{5} ) ^{7} ) ^{\frac{3}{4}}}

Solution steps

Apply the chain rule:

\frac{1}{4 ( ( 2 x + x ^{5} ) ^{7} ) ^{\frac{3}{4}}} \frac{d}{d x} ((2 x + x^{5})^{7})

= \frac{1}{4 ( ( 2 x + x ^{5} ) ^{7} ) ^{\frac{3}{4}}} \frac{d}{d x} ((2 x + x^{5})^{7})

\frac{d}{d x} ((2 x + x^{5})^{7}) = 7 (2 x + x^{5})^{6} (2 + 5 x^{4})

= \frac{1}{4 ( ( 2 x + x ^{5} ) ^{7} ) ^{\frac{3}{4}}} \cdot 7 (2 x + x^{5})^{6} (2 + 5 x^{4})

Simplify

\frac{1}{4 ( ( 2 x + x ^{5} ) ^{7} ) ^{\frac{3}{4}}} \cdot 7 (2 x + x^{5})^{6} (2 + 5 x^{4}) : \frac{7 ( 2 x + x ^{5} ) ^{6} ( 2 + 5 x ^{4} )}{4 ( ( 2 x + x ^{5} ) ^{7} ) ^{\frac{3}{4}}}

= \frac{7 ( 2 x + x ^{5} ) ^{6} ( 2 + 5 x ^{4} )}{4 ( ( 2 x + x ^{5} ) ^{7} ) ^{\frac{3}{4}}}

y^{^{'}} = \frac{( x + 3 y )}{( y - 3 x )}

sec (5 x)

f (x) = 5 x^{2} + 5 x + 7

\frac{\partial}{\partial x} (arctan (\frac{y}{( x - 1 )}))

\frac{\partial}{\partial x} (62 - 2 x^{2} - 2 y^{2})

What is the derivative of \sqrt[4]{(2x+x^5)^7} ?
The derivative of \sqrt[4]{(2x+x^5)^7} is (7(2x+x^5)^6(2+5x^4))/(4((2x+x^5)^7)^{3/4)}
What is the first derivative of \sqrt[4]{(2x+x^5)^7} ?
The first derivative of \sqrt[4]{(2x+x^5)^7} is (7(2x+x^5)^6(2+5x^4))/(4((2x+x^5)^7)^{3/4)}