Question

Use the the limit definition to compute the derivative of the function f(x)=5^x−2 at x=1, and find an equation of the tangent line.

f′(1)=

At x=1, the tangent line is y=

Answer 1

\[
f'(a) = \lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h} \\
f'(1) = \lim_{h \rightarrow 0}\frac{f(1+h)-f(1)}{h} =
\lim_{h \rightarrow 0}\frac{(5^{1+h}-2)-(5^1-2)}{h} = ?\\
\]

Answer 2

\[
f'(1) = \lim_{h \rightarrow 0}\frac{(5^{1+h}-2)-(5^1-2)}{h} =
\lim_{h \rightarrow 0}\frac{((5*5^h-2)-(3))}{h} = \\
\lim_{h \rightarrow 0}\frac{5*5^h-5}{h} = \lim_{h \rightarrow 0}\frac{5(5^h-1)}{h} =
5\lim_{h \rightarrow 0}\frac{(5^h-1)}{h} \\
\text{Let } 1/t = 5^h-1 \\
h \rightarrow 0, ~~ t \rightarrow \infty\\
1+1/t = 5^h \\
\ln(1+1/t) = h\ln(5) \\
h = \ln(1+1/t) / \ln(5) \\
5\lim_{h \rightarrow 0}\frac{(5^h-1)}{h} =
5\lim_{t \rightarrow \infty}\frac{(1/t)}{\ln(1+1/t)/\ln(5)} =
5\ln(5)\lim_{t \rightarrow \infty}\frac{1}{t\ln(1+1/t)} = \\
5\ln(5)\lim_{t \rightarrow \infty}\frac{1}{\ln(1+1/t)^t} =
5\ln(5)\frac{1}{\ln(e)} = 5\ln(5)\\

\\

\\
\]

Answer 3

when \(x = 1,~~ f(x) = 5^1-2=3\)
(1,3) is a point on the curve.
\(y - y_1 = m(x - x_1) \\
y - 3 = 5\ln(5)(x - 1) \\
y = 5\ln(5)x + (3 - 5\ln(5))
\)
is the equation of the tangent at x = 1.