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#.
