How do you find the derivative of #y=(e^(5x^4))/(e^(4x^2+3))#?

Question

Emily Ammerman · Accepted Answer

#dy/dx = (20x^3 - 8x)e^(5x^4 - 4x^2 - 3)#

Use the exponent law #x^a/x^n = x^(a - n)# to write as a single exponent.

#y= e^(5x^4 - (4x^2 + 3))#

#y = e^(5x^4 - 4x^2 - 3)#

Differentiate this using the chain rule. Let #y = e^u# and #u = 5x^4 - 4x^2 - 3#. We know that #d/dx(e^x) = e^x# and that #(du)/dx = 20x^3 - 8x#. Hence:

#dy/dx = dy/(du) * (du)/dx#

#dy/dx = e^u * 20x^3 - 8x#

#dy/dx = (20x^3 - 8x)e^(5x^4 - 4x^2 - 3)#

Hopefully this helps!

Ezekiel Alexander · Answer

undefined To find the derivative of $ y = \frac{e^{5x^4}}{e^{4x^2+3}} $, you can use the quotient rule.

The quotient rule states that if you have a function $ y = \frac{u}{v} $, then its derivative is given by $ y' = \frac{u'v - uv'}{v^2} $.

First, find the derivatives of $ u $ and $ v $ separately. Then apply the quotient rule to find the derivative of $ y $.

Let $ u = e^{5x^4} $ and $ v = e^{4x^2+3} $.

$ u' = \frac{d}{dx}(e^{5x^4}) $ and $ v' = \frac{d}{dx}(e^{4x^2+3}) $.

Using the chain rule, $ u' = 20x^3 e^{5x^4} $ and $ v' = (8x)e^{4x^2+3} $.

Now, apply the quotient rule:

$$ y' = \frac{(20x^3 e^{5x^4})(e^{4x^2+3}) - (e^{5x^4})(8x)e^{4x^2+3}}{(e^{4x^2+3})^2} $$

Simplify this expression to get the derivative of $ y $.