What is the linear approximation to ln((1+𝑎𝑥)^𝑟) at 𝑥=0 ?
If y = ln(1+ax)^r = r ln(1+ax)
y' = ra/(1+ax)
y(0) = 0
y'(0) = ra
so, the tangent line at (0,0) is
y = rax
To find the linear approximation to the function ln((1+𝑎𝑥)^𝑟) at 𝑥=0, we can use the first-order Taylor expansion or linear approximation.
The first step is to find the derivatives of the function.
The derivative of ln((1+𝑎𝑥)^𝑟) with respect to 𝑥 can be found using the chain rule.
Let's denote the function as f(𝑥) = ln((1+𝑎𝑥)^𝑟).
Using the chain rule, the derivative of f(𝑥) with respect to 𝑥 is:
f'(𝑥) = r(1+𝑎𝑥)^(𝑟-1) *𝑎
At 𝑥=0, this derivative simplifies to:
f'(0) = r(1+0)^(𝑟-1) *𝑎
= r * 𝑎
Now, we use the linear approximation formula:
𝑓(𝑥) ≈ 𝑓(0) + 𝑓'(0) * 𝑥
Substituting the values in, we have:
ln((1+𝑎𝑥)^𝑟) ≈ ln((1+0)^𝑟) + (r * 𝑎) * 𝑥
≈ ln(1) + (r * 𝑎) * 𝑥
≈ 0 + (r * 𝑎) * 𝑥
≈ r * 𝑎 * 𝑥
Therefore, the linear approximation to ln((1+𝑎𝑥)^𝑟) at 𝑥=0 is r * 𝑎 * 𝑥.
To find the linear approximation of the function ln((1+𝑎𝑥)^𝑟) at 𝑥=0, we can use the first two terms of the Taylor series expansion. The Taylor series expansion for a function f(x) centered at 𝑥=0 is given by:
f(x) = f(0) + f'(0)x + ...
In this case, we need to find the first term (f(0)) and the coefficient of the linear term (f'(0)).
First, we find the value of the function at 𝑥=0 by plugging in 𝑥=0 into the given function ln((1+𝑎𝑥)^𝑟). The result is ln((1+0)^𝑟) = ln(1) = 0. So, f(0) = 0.
Next, we need to find the derivative of the function ln((1+𝑎𝑥)^𝑟) with respect to 𝑥, and evaluate it at 𝑥=0. Let's differentiate the function using the chain rule:
f'(x) = 𝑟(1+𝑎𝑥)^(𝑟-1) * a = 𝑟𝑎(1+𝑎𝑥)^(𝑟-1)
Now we can evaluate this derivative at 𝑥=0:
f'(0) = 𝑟𝑎(1+𝑎(0))^(𝑟-1) = 𝑟𝑎(1)^𝑟-1 = 𝑟𝑎
Now that we have found f(0) = 0 and f'(0) = 𝑟𝑎, we can write the linear approximation:
ln((1+𝑎𝑥)^𝑟) ≈ f(0) + f'(0)x = 0 + 𝑟𝑎x = 𝑟𝑎x
Therefore, the linear approximation to ln((1+𝑎𝑥)^𝑟) at 𝑥=0 is 𝑟𝑎x.