find the derivative
ln [x^3 +((x+3)^3)((x^2)+4)^7
3 answers
You left out a ] somewhere
Hmm the logical fix would be
ln [x^3 +((x+3)^3)]((x^2)+4)^7
a^3 + b^3 = (a+b)(a^2-ab+b^2), so
x^3 + (x+3)^3 = (x+(x+3))(x^2 - x(x+3) + (x+3)^2)
= (2x+3)(x^2 + 3x + 9)
and the log of the product then becomes
y = ln(2x+3) + ln(x^2+3x+9) + 7ln(x^2+4)
y' = 2/(2x+3) + (2x+3)/(x^2+3x+9) + 14x/(x^2+4)
If my placement of [] is wrong, please feel free to clarify
ln [x^3 +((x+3)^3)]((x^2)+4)^7
a^3 + b^3 = (a+b)(a^2-ab+b^2), so
x^3 + (x+3)^3 = (x+(x+3))(x^2 - x(x+3) + (x+3)^2)
= (2x+3)(x^2 + 3x + 9)
and the log of the product then becomes
y = ln(2x+3) + ln(x^2+3x+9) + 7ln(x^2+4)
y' = 2/(2x+3) + (2x+3)/(x^2+3x+9) + 14x/(x^2+4)
If my placement of [] is wrong, please feel free to clarify
Find the derivative of the function
y=ln(2x/x+3)
y=ln(2x/x+3)
1 answer