What is the derivative of f(x)=ln(19x^3+sin(3x^2)) ?

The derivative of f(x)=ln(19x^3+sin(3x^2)) is (57x^2+6xcos(3x^2))/(19x^3+sin(3x^2))

What is the first derivative of f(x)=ln(19x^3+sin(3x^2)) ?

The first derivative of f(x)=ln(19x^3+sin(3x^2)) is (57x^2+6xcos(3x^2))/(19x^3+sin(3x^2))

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derivative of

f (x) = ln (19 x^{3} + sin (3 x^{2}))

\frac{57 x ^{2} + 6 x cos ( 3 x ^{2} )}{19 x ^{3} + sin ( 3 x ^{2} )}

Solution steps

\frac{d}{d x} (ln (19 x^{3} + sin (3 x^{2})))

Apply the chain rule:

\frac{1}{19 x ^{3} + sin ( 3 x ^{2} )} \frac{d}{d x} (19 x^{3} + sin (3 x^{2}))

= \frac{1}{19 x ^{3} + sin ( 3 x ^{2} )} \frac{d}{d x} (19 x^{3} + sin (3 x^{2}))

\frac{d}{d x} (19 x^{3} + sin (3 x^{2})) = 57 x^{2} + cos (3 x^{2}) \cdot 6 x

= \frac{1}{19 x ^{3} + sin ( 3 x ^{2} )} (57 x^{2} + cos (3 x^{2}) \cdot 6 x)

Simplify

\frac{1}{19 x ^{3} + sin ( 3 x ^{2} )} (57 x^{2} + cos (3 x^{2}) \cdot 6 x) : \frac{57 x ^{2} + 6 x cos ( 3 x ^{2} )}{19 x ^{3} + sin ( 3 x ^{2} )}

= \frac{57 x ^{2} + 6 x cos ( 3 x ^{2} )}{19 x ^{3} + sin ( 3 x ^{2} )}

\frac{d}{d X} (X^{3} - 3 X)

x y^{^{'}} + y = - 8 x y^{2}, y (1) = 3

\frac{s + 1}{s ^{2} + 4 s + 8}

sin ((π x)^{8})

\int x sin (3 x^{2} + 10) d x

What is the derivative of f(x)=ln(19x^3+sin(3x^2)) ?
The derivative of f(x)=ln(19x^3+sin(3x^2)) is (57x^2+6xcos(3x^2))/(19x^3+sin(3x^2))
What is the first derivative of f(x)=ln(19x^3+sin(3x^2)) ?
The first derivative of f(x)=ln(19x^3+sin(3x^2)) is (57x^2+6xcos(3x^2))/(19x^3+sin(3x^2))