Given f(x)= 3e^(1-x^2)*ln(x), find the equation of the tangent line at x = 1.
f'(x) = -3/x e^(1-x^2) (2x^2 lnx - 1)
f(1) = 0
f'(1) = -3(1)(-1) = 3
So, now you have a point (1,0) and a slope (3) and the line is
y-0 = 3(x-1)
To find the equation of the tangent line at x = 1, we need to find the slope of the tangent (m) and the point of tangency (x, y) on the curve f(x) = 3e^(1-x^2) * ln(x).
Let's start by finding the slope of the tangent line.
The slope of the tangent line (m) can be found using the derivative of the function f(x).
f'(x) is the derivative of f(x). Let's find it first:
f'(x) = d/dx [3e^(1-x^2) * ln(x)]
To differentiate this product of functions, we can use the product rule. The product rule states that:
d/dx[f(x) * g(x)] = f'(x) * g(x) + f(x) * g'(x)
Let's apply the product rule to differentiate f(x)= 3e^(1-x^2) * ln(x):
f(x) = 3e^(1-x^2)
g(x) = ln(x)
f'(x) = (3)(d/dx[e^(1-x^2)]) * ln(x) + (3e^(1-x^2)) * (d/dx[ln(x)])
To differentiate e^(1-x^2), we can use the chain rule:
d/dx[e^(u)] = u' * e^(u)
Let u = 1-x^2
u' = d/dx(1-x^2) = -2x
Now, let's differentiate ln(x):
d/dx[ln(x)] = 1/x
Now we can substitute these derivatives back into f'(x):
f'(x) = (3)(-2x)(e^(1-x^2)) * ln(x) + (3e^(1-x^2)) * (1/x)
Simplifying further, we get:
f'(x) = -6x e^(1-x^2) ln(x) + 3e^(1-x^2)/x
Next, let's find the slope of the tangent line at x = 1 by plugging x = 1 into f'(x):
m = f'(1) = -6(1) e^(1-(1^2)) ln(1) + 3e^(1-(1^2))/1
= -6e^(0) * 0 + 3e^(0)
= 3
So the slope of the tangent line at x = 1 is 3.
Now let's find the corresponding y-coordinate on the curve by plugging x = 1 into f(x):
f(1) = 3e^(1-(1^2)) ln(1)
= 3e^0 * ln(1)
= 3 * 0
= 0
The point of tangency (1, y) on the curve is (1, 0).
Thus, we have the slope of the tangent (m = 3) and the point of tangency (1, 0).
Finally, we can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values we found, we get:
y - 0 = 3(x - 1)
Simplifying further:
y = 3x - 3
Therefore, the equation of the tangent line at x = 1 is y = 3x - 3.
To find the equation of the tangent line at x = 1 for the function f(x) = 3e^(1-x^2) * ln(x), follow these steps:
Step 1: Find the derivative of the function f(x). The derivative of f(x) will give us the slope of the tangent line at any point on the graph of f(x).
Step 2: Evaluate the derivative at x = 1 to find the slope of the tangent line at that particular point.
Step 3: Use the point-slope form of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line, to find the equation of the tangent line.
Let's go through each step in detail:
Step 1: Find the derivative of f(x).
The function f(x) = 3e^(1-x^2) * ln(x) is a product of two functions, so we can use the product rule to find its derivative.
Using the product rule, the derivative of f(x) is:
f'(x) = (3e^(1-x^2) * d/dx(ln(x))) + ((d/dx(3e^(1-x^2))) * ln(x))
The derivative of ln(x) is 1/x.
The derivative of e^u, where u is a function of x, is (d/dx)e^u = (d/du)e^u * (du/dx).
Applying the product rule to the first term:
f'(x) = (3e^(1-x^2) * 1/x) + ((d/dx(3e^(1-x^2))) * ln(x))
Now, let's find the derivative of the second term:
The derivative of 3e^(1-x^2) is (d/dx(3e^(1-x^2))) = 3 * (d/dx(e^(1-x^2))).
Using the chain rule, the derivative of e^(1-x^2) is (d/dx(e^(1-x^2))) = -(2x)e^(1-x^2).
Substituting this into the derivative:
f'(x) = (3e^(1-x^2) * 1/x) + (3 * (-(2x)e^(1-x^2)) * ln(x))
Simplifying further:
f'(x) = 3e^(1-x^2)/x - 6x*e^(1-x^2) * ln(x)
Step 2: Evaluate the derivative at x = 1 to find the slope of the tangent line at that point.
To find the slope of the tangent line at x = 1, substitute x = 1 into the derivative:
f'(1) = 3e^(1 - (1^2))/1 - 6(1)e^(1 - (1^2)) * ln(1)
Simplifying:
f'(1) = 3e^0 - 6(1)e^0 * ln(1)
= 3 - 6(0) * ln(1)
= 3
So, the slope of the tangent line at x = 1 is 3.
Step 3: Use the point-slope form to find the equation of the tangent line.
We know that x = 1 is a point on the tangent line with a slope of 3. We can use the point-slope form, y - y1 = m(x - x1), where (x1, y1) is the point (1, f(1)) on the graph of f(x).
To find y1, substitute x = 1 into f(x):
f(1) = 3e^(1 - (1^2)) * ln(1)
= 3e^0 * ln(1)
= 3 * 0
= 0
So, the point (1, 0) lies on the graph of f(x).
Now we can write the equation of the tangent line using the point (1, 0) and the slope found in Step 2:
y - 0 = 3(x - 1)
y = 3x - 3
Therefore, the equation of the tangent line at x = 1 is y = 3x - 3.