What is (d^4)/(dx^4)(e^{-4x})+6ln(3x^5) ?

The solution to (d^4)/(dx^4)(e^{-4x})+6ln(3x^5) is (d^4)/(dx^4)(e^{-4x})+6ln(3)+30ln(x)

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simplify (d^4)/(dx^4)(e^{-4x})+6ln(3x^5)

simplify

\frac{d ^{4}}{d x ^{4}} (e^{- 4 x}) + 6 ln (3 x^{5})

\frac{d ^{4}}{d x ^{4}} (e^{- 4 x}) + 6 ln (3) + 30 ln (x)

Solution steps

\frac{d ^{4}}{d x ^{4}} (e^{- 4 x}) + 6 ln (3 x^{5})

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

6 ln (3 x^{5}) = 6 ln (3) + 6 ln (x^{5})

= \frac{d ^{4}}{d x ^{4}} (e^{- 4 x}) + 6 ln (3) + 6 ln (x^{5})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (x^{5}) = 5 ln (x)

= \frac{d ^{4}}{d x ^{4}} (e^{- 4 x}) + 6 ln (3) + 6 \cdot 5 ln (x)

Multiply the numbers:

6 \cdot 5 = 30

= \frac{d ^{4}}{d x ^{4}} (e^{- 4 x}) + 6 ln (3) + 30 ln (x)

ln (y) - \frac{1}{2} ln (y - 1) - \frac{1}{2} ln (y + 1)

ln (\frac{\frac{1}{2}}{\frac{1}{4}})

- 2 ln ∣ l + d ∣ + ln ∣ d ∣ + ln ∣ 2 l + d ∣

lo g_{4} (x^{2} - 1) - 5 lo g_{4} (x - 1)

ln (1) - ln (\frac{1}{3})

What is (d^4)/(dx^4)(e^{-4x})+6ln(3x^5) ?
The solution to (d^4)/(dx^4)(e^{-4x})+6ln(3x^5) is (d^4)/(dx^4)(e^{-4x})+6ln(3)+30ln(x)