What is the derivative of ln(3x^2-cos(x)) ?

The derivative of ln(3x^2-cos(x)) is (6x+sin(x))/(3x^2-cos(x))

What is the first derivative of ln(3x^2-cos(x)) ?

The first derivative of ln(3x^2-cos(x)) is (6x+sin(x))/(3x^2-cos(x))

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\frac{d}{d x} (ln (3 x^{2} - cos (x)))

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

Solution steps

\frac{d}{d x} (ln (3 x^{2} - cos (x)))

Apply the chain rule:

\frac{1}{3 x ^{2} - cos ( x )} \frac{d}{d x} (3 x^{2} - cos (x))

= \frac{1}{3 x ^{2} - cos ( x )} \frac{d}{d x} (3 x^{2} - cos (x))

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

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

Simplify

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

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

y^{^{'}} - y tan (x) = sec (x)

(3 x - 3)^{2} (1 - 5 x)^{4}

3 t + 2

y^{^{''}} + 2 y + y = 0

f (x) = (\frac{5}{x ^{2} + 1}), at x = - 1

What is the derivative of ln(3x^2-cos(x)) ?
The derivative of ln(3x^2-cos(x)) is (6x+sin(x))/(3x^2-cos(x))
What is the first derivative of ln(3x^2-cos(x)) ?
The first derivative of ln(3x^2-cos(x)) is (6x+sin(x))/(3x^2-cos(x))