Tolong sertakan cara pengerjaannya

Tolong sertakan cara pengerjaannya

$displaystyle y = frac{ln(x+3)}{ln(2-x)} \ frac{dy}{dx} = y' = dots?$

Penyelesaian:

• Turunan fungsi bentuk $y =frac{u}{v}$

$displaystyle boxed{y' = frac{vu'-uv'}{v^2}}$

• Turunan fungsi bentuk $y = lnbig(f(x)big)$

$displaystyle boxed{y' = frac{1}{f(x)}}$

Dari soal di atas, misal:

$displaystyle begin{array}{ll}u = ln(x+3) & v = ln(2-x) \ u' = frac{1}{x+3} & v' = -frac{1}{2-x} end{array}$

sehingga

$displaystyle y = frac{u}{v} \ begin{aligned} y' &= frac{vu'-uv'}{v^2} \ &= frac{ln(2-x)cdotfrac{1}{x+3}-ln(x+3)cdotleft( -frac{1}{2-x} right)}{left(ln(2-x)right)^2} \ &= frac{frac{ln(2-x)}{x+3}+frac{ln(x+3)}{2-x}}{ln^2(2-x)} \ &= frac{(2-x)ln(2-x)+(x+3)ln(x+3)}{(x+3)(2-x)ln^2(2-x)}end{aligned}\$

Jawaban:

$displaystyle boxed{bold{frac{(2-x)ln(2-x)+(x+3)ln(x+3)}{(x+3)(2-x)ln^2(2-x)} }}$