What is the derivative of ln(2x^4e^{3x}(x^2+x+1^5)) ?

The derivative of ln(2x^4e^{3x}(x^2+x+1^5)) is (14x^2+9x+4)/(x(x^2+x+1))+3

What is the first derivative of ln(2x^4e^{3x}(x^2+x+1^5)) ?

The first derivative of ln(2x^4e^{3x}(x^2+x+1^5)) is (14x^2+9x+4)/(x(x^2+x+1))+3

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

\frac{14 x ^{2} + 9 x + 4}{x ( x ^{2} + x + 1 )} + 3

Solution steps

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

Simplify

ln (2 x^{4} e^{3 x} (x^{2} + x + 1)^{5}) : ln (2 x^{4} (x^{2} + x + 1)^{5}) + 3 x

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (ln (2 x^{4} (x^{2} + x + 1)^{5})) = \frac{14 x ^{2} + 9 x + 4}{x ( x ^{2} + x + 1 )}

\frac{d}{d x} (3 x) = 3

= \frac{14 x ^{2} + 9 x + 4}{x ( x ^{2} + x + 1 )} + 3

\int \frac{3 - x}{3 - x ^{2}} d x

\int_{- 2}^{5} - 2 d x

\int \frac{x - 2}{( x ^{2} + x + 2 ) ( x ^{2} + 2 )} d x

\int_{0}^{4} π (2 x)^{2} d x

\frac{d}{d x} (x ln (x) - x + 1)

What is the derivative of ln(2x^4e^{3x}(x^2+x+1^5)) ?
The derivative of ln(2x^4e^{3x}(x^2+x+1^5)) is (14x^2+9x+4)/(x(x^2+x+1))+3
What is the first derivative of ln(2x^4e^{3x}(x^2+x+1^5)) ?
The first derivative of ln(2x^4e^{3x}(x^2+x+1^5)) is (14x^2+9x+4)/(x(x^2+x+1))+3