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How do I find the derivative of #F(y) = yln(9 + e^y)#?

Answer 1

#f'(y)=ye^y/(9+e^y)+ln(9+e^y)#

First, turn it into #f(x)=xln(9+e^x)# (It is not necessary but x is less confusing in the equation) Next, apply product rule #f'(x)=x*(1/(9+e^x) * e^x)+1*(ln(9+e^x))# Note the chain rule used when taking derivative of #ln(9+e^x)# Simplify #f'(x)=xe^x/(9+e^x)+ln(9+e^x)# Lastly, replace all x with y

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Answer 2

Evelyn Agard

To find the derivative of F(y) = yln(9 + e^y), you can use the product rule and the chain rule of differentiation.

Apply the product rule: (uv)' = u'v + uv'
Let u = y and v = ln(9 + e^y)
Find the derivatives of u and v.
- u' = 1
- v' = (1/(9 + e^y)) * (e^y)
Substitute u', v', u, and v into the product rule formula.
- F'(y) = (1)(ln(9 + e^y)) + (y)((1/(9 + e^y)) * (e^y))
Simplify the expression.
- F'(y) = ln(9 + e^y) + (y * e^y) / (9 + e^y)

So, the derivative of F(y) = yln(9 + e^y) is F'(y) = ln(9 + e^y) + (y * e^y) / (9 + e^y).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.