First, find the derivative of this function,
f(x) = x^3 ln(x^2 + 1) - 2x.
Using the sum rule, we could split this into two parts,
g(x) = x^3 ln(x^2 + 1)
and
h(x) = -2x,
so that
f(x) = g(x) + h(x)
and
(df)/(dx) = (dg)/(dx) + (dh)/(dx).
We could then proceed to find the derivative of each function, starting with g(x):
(dg)/(dx) = d/dx (x^3 ln(x^2 + 1))
Using the product rule:
dg = x^3 * d(ln(x^2 + 1)) + ln(x^2 + 1) * d(x^3)
Let's solve for each differential, from derivatives, starting from d(x^3):
(d(x^3))/(dx) = 3x^2 rarr d(x^3) = 3x^2 dx
Putting that back in:
dg = x^3 * d(ln(x^2 + 1)) + ln(x^2 + 1) * 3x^2 dx
Now, how about the other differential? I think solving for that derivative requires breaking it up into smaller functions. Let's set
p(x) = ln(x^2 + 1) = p_3(p_2(p_1(x)))
where
p_1(x) = x^2
p_2(x) = x + 1
p_3(x) = ln(x)
then use the chain rule:
(dp)/(dx) = (dp_1)/(dx) * (dp_2)/(dp_1) * (dp_3)/(dp_2)
Starting from the derivative of p_1 to x:
(dp_1)/(dx) = (d(x^2))/(dx) = 2x
Then solving for the differential, by "multiplying" by dx:
(dp_1)/(dx) = 2x rarr dp_1 = 2x dx
Next, p_2 to p_1:
(dp_2)/(dp_1) = (d(p_1 + 1))/(dp_1) = 1
(dp_2)/(dp_1) = 1 rarr dp_2 = dp_1
Awesome! Evaluate dp_1:
dp_2 = 2x dx
Then dp_3 to dp_2:
(dp_3)/(dp_2) = (d(ln(p_2)))/(dp_2) = 1/(p_2)
(dp_3)/(dp_2) = 1/(p_2) rarr dp_3 = 1/(p_2) dp_2
dp_3 = 1/(p_1 + 1) 2x dx = (2x)/(x^2 + 1) dx
So:
(dp_3)/(dx) = (dp)/(dx) = (2x)/(x^2 + 1)
(dp)/(dx) = (2x)/(x^2 + 1) rarr dp = (2x)/(x^2 + 1) dx
Substituting dp back in dg:
dg = x^3 * dp + ln(x^2 + 1) * 3x^2 dx
dg = x^3 * (2x)/(x^2 + 1) dx + ln(x^2 + 1) * 3x^2 dx
Simplifying:
dg = (2x^4)/(x^2 + 1) dx + 3x^2 ln(x^2 + 1) dx
Dividing by dx to solve for the derivative:
(dg)/(dx) = (2x^4)/(x^2 + 1) + 3x^2 ln(x^2 + 1)
And substituting this back into (df)/(dx):
(df)/(dx) = (2x^4)/(x^2 + 1) + 3x^2 ln(x^2 + 1) + (dh)/(dx)
Now, on to h(x) = -2x:
(dh)/(dx) = -2
So:
(df)/(dx) = (2x^4)/(x^2 + 1) + 3x^2 ln(x^2 + 1) - 2
Let's solve for x = 1:
rarr (2(1)^4)/((1)^2 + 1) + 3(1)^2 ln((1)^2 + 1) - 2
= 1 + 3 ln(2) - 2
m_t = 3 ln(2) - 1
That gives us the slope of the tangent line. Taking the negative reciprocal:
m_n = -(m_t)^(-1) = -(3 ln(2) - 1)^(-1) = 1/(1 - 3 ln(2))
Now that we have the slope of the normal line, we just need the y-intercept. We can start by using the point-slope form:
y - y_1 = m(x - x_1)
The slope, m, in this case is m_n = 1/(1 - 3 ln(2)) :
y - y_1 = (x - x_1)/(1 - 3 ln(2))
We need a point on the graph. Well, there's our input, x = 1, and the output, f(1) = (1)^3 ln((1)^2 + 1) - 2(1) = ln(2) - 2, so substitute that for x_1 and y_1 respectively:
y - (ln(2) - 2) = (x - 1)/(1 - 3 ln(2))
Finally, isolate y:
y = (x - 1)/(1 - 3 ln(2)) + ln(2) - 2
That's the equation for the line normal to f(x) = x^3 ln(x^2 + 1) - 2x at x = 1.
