What is the derivative of f(x)=\sqrt[x^4]{7-9x} ?

The derivative of f(x)=\sqrt[x^4]{7-9x} is (-(4ln(7-9x))/(x^5)-9/(x^4(7-9x)))(7-9x)^{1/(x^4)}

What is the first derivative of f(x)=\sqrt[x^4]{7-9x} ?

The first derivative of f(x)=\sqrt[x^4]{7-9x} is (-(4ln(7-9x))/(x^5)-9/(x^4(7-9x)))(7-9x)^{1/(x^4)}

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derivative of

(- \frac{4 ln ( 7 - 9 x )}{x ^{5}} - \frac{9}{x ^{4} ( 7 - 9 x )}) (7 - 9 x)^{\frac{1}{x ^{4}}}

Solution steps

Apply exponent rule:

a^{b} = e^{b l n (a)}

= \frac{d}{d x} (e^{\frac{1}{x ^{4}} l n (7 - 9 x)})

Apply the chain rule:

e^{\frac{1}{x ^{4}} l n (7 - 9 x)} \frac{d}{d x} (\frac{1}{x ^{4}} ln (7 - 9 x))

= e^{\frac{1}{x ^{4}} l n (7 - 9 x)} \frac{d}{d x} (\frac{1}{x ^{4}} ln (7 - 9 x))

\frac{d}{d x} (\frac{1}{x ^{4}} ln (7 - 9 x)) = - \frac{4 ln ( 7 - 9 x )}{x ^{5}} - \frac{9}{x ^{4} ( 7 - 9 x )}

= e^{\frac{1}{x ^{4}} l n (7 - 9 x)} (- \frac{4 ln ( 7 - 9 x )}{x ^{5}} - \frac{9}{x ^{4} ( 7 - 9 x )})

e^{\frac{1}{x ^{4}} l n (7 - 9 x)} = (7 - 9 x)^{\frac{1}{x ^{4}}}

= (- \frac{4 ln ( 7 - 9 x )}{x ^{5}} - \frac{9}{x ^{4} ( 7 - 9 x )}) (7 - 9 x)^{\frac{1}{x ^{4}}}

\int_{0}^{1} 4^{4 x} d x

\int \frac{cos ( x )}{sin ^{2} ( x ) - 64} d x

\int cos (2 π) d x

\frac{d}{d x} (\frac{cos ( 6 x )}{e ^{x}})

y = arctan (x)

What is the derivative of f(x)=\sqrt[x^4]{7-9x} ?
The derivative of f(x)=\sqrt[x^4]{7-9x} is (-(4ln(7-9x))/(x^5)-9/(x^4(7-9x)))(7-9x)^{1/(x^4)}
What is the first derivative of f(x)=\sqrt[x^4]{7-9x} ?
The first derivative of f(x)=\sqrt[x^4]{7-9x} is (-(4ln(7-9x))/(x^5)-9/(x^4(7-9x)))(7-9x)^{1/(x^4)}