Given f(x)=6^5-x^2 find f'(2)

Question

I really need to learn how to do this problem. Thanks so much!

Answer 1

Your equation is ambiguous as written.  I'm not sure if it's f(x) = 6⁵ - x² or f(x) = 6^{5-x^2}.  I'll do both.

 

f(x) = 6⁵ - x²

f'(x) = df(x)/dx = d(6⁵)/dx - d(x²)/dx = -2x

f'(2) = -2(2) = -4

 

f(x) = 6^{5-x^2}

Rewrite as f(x) = e^{ln(6^(5-x^2))} = e^{(5-x^2)·ln(6)}         [LOG PROPERTY: ln(a^b) =b·ln(a)]

Apply the chain rule, letting u = 5-x2

f'(x) = df(x)/dx = df(u).du · du/dx

f'(x) = d(e^ln6·u)/du · d(5-x²)/dx

f'(x) = ln(6)e^u · (-2x)

f'(x) = ln(6)e^{(5-x^2)}·(-2x)

f'(2)( = -4·ln(6)·e

Use your calculator to get the answer.

Answer 2

If you evaluate the limit

 

lim     [f(2 + h) - f(2)] / h

x→0 

 

 

You should get the value of -4, as Arturo obtained.  But this approach if for those who have not yet learned the derivative rules.

 

Answer 3

f(x) = -x² + 6⁵

 

Just use the power rule.  Note that 6⁵ is a constant.

 

f'(x) = -2x

 

f'(2) = -2(2) = -4