How do you differentiate $y = \frac{e^{5 x^{4}}}{e^{4 x^{2} + 3}}$ $y=(e^(5x^4))/(e^(4x^2+3))$ ?

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Answer 1 · 2016-11-29T18:19:16

Rewrite as $y = e^{5 x^{4} - 4 x^{2} - 3}$ $y = e^(5x^4-4x^2-3)$ , then use $\frac{d}{d x} (e^{u}) = e^{u} \frac{d u}{d x}$ $d/dx(e^u) = e^u (du)/dx$

An important property of exponents is $\frac{a^{n}}{a^{m}} = a^{n - m}$ $a^n/a^m = a^(n-m)$

$y = \frac{e^{5 x^{4}}}{e^{4 x^{2} + 3}} = e^{5 x^{4} - 4 x^{2} - 3}$ $y = e^(5x^4)/e^(4x^2+3) = e^(5x^4-4x^2-3)$

$y' = e^(5x^4-4x^2-3) * d/dx(5x^4-4x^2-3)$

$= (20x^3-8x)e^(5x^4-4x^2-3)$

How do you differentiate y=(e^(5x^4))/(e^(4x^2+3))y=e5x4e4x2+3y=(e^(5x^4))/(e^(4x^2+3))?