What is the derivative of log_{4}(sec^2(5x^3-4x^2)) ?

The derivative of log_{4}(sec^2(5x^3-4x^2)) is (tan(5x^3-4x^2)(15x^2-8x))/(ln(2))

What is the first derivative of log_{4}(sec^2(5x^3-4x^2)) ?

The first derivative of log_{4}(sec^2(5x^3-4x^2)) is (tan(5x^3-4x^2)(15x^2-8x))/(ln(2))

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derivative of

lo g_{4} (sec^{2} (5 x^{3} - 4 x^{2}))

\frac{tan ( 5 x ^{3} - 4 x ^{2} ) ( 15 x ^{2} - 8 x )}{ln ( 2 )}

Solution steps

\frac{d}{d x} (lo g_{4} (sec^{2} (5 x^{3} - 4 x^{2})))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( sec ^{2} ( 5 x ^{3} - 4 x ^{2} ) )}{ln ( 4 )})

Take the constant out:

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

= \frac{1}{ln ( 4 )} \frac{d}{d x} (ln (sec^{2} (5 x^{3} - 4 x^{2})))

Apply the chain rule:

\frac{1}{sec ^{2} ( 5 x ^{3} - 4 x ^{2} )} \frac{d}{d x} (sec^{2} (5 x^{3} - 4 x^{2}))

= \frac{1}{sec ^{2} ( 5 x ^{3} - 4 x ^{2} )} \frac{d}{d x} (sec^{2} (5 x^{3} - 4 x^{2}))

\frac{d}{d x} (sec^{2} (5 x^{3} - 4 x^{2})) = 2 sec^{2} (5 x^{3} - 4 x^{2}) tan (5 x^{3} - 4 x^{2}) (15 x^{2} - 8 x)

= \frac{1}{ln ( 4 )} \cdot \frac{1}{sec ^{2} ( 5 x ^{3} - 4 x ^{2} )} \cdot 2 sec^{2} (5 x^{3} - 4 x^{2}) tan (5 x^{3} - 4 x^{2}) (15 x^{2} - 8 x)

Simplify

\frac{1}{ln ( 4 )} \cdot \frac{1}{sec ^{2} ( 5 x ^{3} - 4 x ^{2} )} \cdot 2 sec^{2} (5 x^{3} - 4 x^{2}) tan (5 x^{3} - 4 x^{2}) (15 x^{2} - 8 x) : \frac{tan ( 5 x ^{3} - 4 x ^{2} ) ( 15 x ^{2} - 8 x )}{ln ( 2 )}

= \frac{tan ( 5 x ^{3} - 4 x ^{2} ) ( 15 x ^{2} - 8 x )}{ln ( 2 )}

\int \frac{2 x - 4 + 3 x}{x} d x

\int_{0}^{30} 500 d x

(y^{3} - 1) e^{x} d x + 3 y^{2} (e^{x} + 1) d y = 0, y (0) = 2

f (x) = 2 (3 x - 8)^{4} - 5 x + 6, at x = 3

\int 9 cos^{5} (x) sin^{4} (x) d x

What is the derivative of log_{4}(sec^2(5x^3-4x^2)) ?
The derivative of log_{4}(sec^2(5x^3-4x^2)) is (tan(5x^3-4x^2)(15x^2-8x))/(ln(2))
What is the first derivative of log_{4}(sec^2(5x^3-4x^2)) ?
The first derivative of log_{4}(sec^2(5x^3-4x^2)) is (tan(5x^3-4x^2)(15x^2-8x))/(ln(2))