Find derivative of the following function in photo attached. Include steps please!?

f(x) = e^(4x² - 3) / Ln(2x + 3) ← this is the function

f(x) = [e^(4x²) * e^(- 3)] / Ln(2x + 3)

f(x) = e^(- 3) * e^(4x²) / Ln(2x + 3)

f(x) = (1/e³) * e^(4x²) / Ln(2x + 3)


This function looks like (1/e³) * (u/v), so its derivative looks like: (1/e³) * [(u'.v) - (v'.u)]/v² → where:

u = e^(4x²) → u' = 8x.e^(4x²)

v = Ln(2x + 3) → v' = 2/(2x + 3)


f'(x) = (1/e³) * [(u'.v) - (v'.u)]/v²

f'(x) = (1/e³) * { [8x.e^(4x²) * Ln(2x + 3)] - [2/(2x + 3)].e^(4x²) } / [Ln(2x + 3)]²

f'(x) = (2/e³) * { [4x.e^(4x²) * Ln(2x + 3)] - [1/(2x + 3)].e^(4x²) } / [Ln(2x + 3)]²

f'(x) = (2/e³) * e^(4x²) * { 4x.Ln(2x + 3) - [1/(2x + 3)] } / [Ln(2x + 3)]²

f'(x) = 2.e^(4x²) * { 4x.Ln(2x + 3) - [1/(2x + 3)] } / e³.[Ln(2x + 3)]²

f'(x) = 2.e^(4x²) * { 4x.(2x + 3).Ln(2x + 3) - 1 } / e³.[Ln(2x + 3)]²

f'(x) = 2.e^(4x²) * [(8x² + 12x).Ln(2x + 3) - 1 } / e³.[Ln(2x + 3)]²

f(x) = e^(4x^2-3) / ln(2x+3)

You may use the quotient rule or the product rule. Let us use the product rule.

f(x) = e^(4x^2-3) (ln (2x+3) )^(-1)
f'(x) = e^(4x^2-3) d/dx ( (ln (2x+3) )^(-1) ) + (ln (2x+3))^(-1) d/dx ( e^(4x^2-3) )

d/dx (ln (2x+3) )^(-1) ) = (-1) (ln (2x+3) )^(-2) d/dx ( ln (2x+3) )
= (-1) (ln (2x+3) )^(-2) ( 2/ (2x+3))
= - 2 / ( (2x+3) (ln (2x+3) )^2 )

d/dx ( e^(4x^2-3) ) = e^(4x^2-3) d/dx (4x^2-3)
= e^(4x^2-3) (8x)
= 8x e^(4x^2-3)

f'(x) = e^(4x^2-3) [ - 2 / ( (2x+3) (ln (2x+3) )^2 ) ] + (ln (2x+3))^(-1) [8x e^(4x^2-3)]
f'(x) = -2 e^(4x^2-3) / ( (2x+3) (ln (2x+3) )^2 ) + 8 x e^(4x^2-3) / ln (2x+3)


8 x e^(4x^2-3) / ln (2x+3) = 8 x e^(4x^2-3)(2x+3) ln(2x+3) / ((2x+3) ln^2 (2x+3))
f'(x) = -2 e^(4x^2-3) / ( (2x+3) (ln (2x+3) )^2 ) + 8 x e^(4x^2-3)(2x+3) ln(2x+3) / ((2x+3) ln^2 (2x+3))
f'(x) = e^(4x^2-3) ( -2 + 8x (2x+3) ln(2x+3) ) / ((2x+3) ln^2 (2x+3))

The photo is not "attached," and the link is not clickable, so it would be a lot of trouble for me to look at it.