# Calc

g'(x)=tan(2/1+x^2)

Let g be the function with first derivative given above and g(1)=5. If f is the function defined by f(x)=ln(g(x)), what is the value of f'(1)?

I know the answer is 0.311, but I need steps as to why. Please and thanks.

1. f = ln g
df/dx = df/dg * dg/dx
we were given dg/dx
what is df/dg ?
df/dg = d ln g / dg = 1/g

so df/dx = (1/g)tan 2/(1+x^2) I think you mean

= (1/5)tan 2/2 = (1/4) tan one radian or 180/pi deg
= (1/5)(1.055741 )
= .31148 so agree

posted by Damon

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