So, we have
#f(x) = 14x^3 - (4x^2 - x)/(e^(3x))#
And want to find the equation of the tangent line at #x = -2#. We do know that the slope of this tangent line would be the derivative of the function:
#(df)/(dx) = d/dx (14x^3 - (4x^2 - x)/(e^(3x)))#
Using the sum rule, we could take the derivative of each term (and take the subtraction out):
#(df)/(dx) = d/dx (14x^3) - d/dx((4x^2 - x)/(e^(3x)))#
First taking the derivative of the term on the left:
#d/dx (14x^3) = 14 d/dx(x^3) = 14 * 3x^2 = 42x^2#
Putting that back in:
#(df)/(dx) = 42x^2 - d/dx((4x^2 - x)/(e^(3x)))#
As for this second function... let's set a function #g(x)# where
#g(x) = (4x^2 - x)/(e^(3x))#
So that
#(df)/(dx) = 42x^2 - (dg)/(dx)#
And we have to take the derivative of #g(x)#. We will first decompose it (break it down) into two smaller functions, one of the numerator, #p(x)#, and the other of the denominator, #q(x)#:
#p(x) = 4x^2 - x#
#q(x) = 1/(e^(3x))#
So that
#g(x) = p(x)q(x)#
And, taking the derivative, by the product rule, we have
#dg = p(x)dq + q(x)dp#
"Dividing" the differentials by #dx#:
#(dg)/(dx) = p(x)(dq)/(dx) + q(x)(dp)/(dx)#
Substituting what we already have:
#(dg)/(dx) = (4x^2 - x)(dq)/(dx) + (1/(e^(3x)))(dp)/(dx)#
Now, we need to take the derivatives of #p(x)# and #q(x)#. Starting from #p(x)#, we can separate its two terms using the sum rule, and take out the subtraction:
#(dp)/(dx) = d/dx (4x^2) - d/dx (x)#
Solve for each term:
#(dp)/(dx) = 4 d/dx (x^2) - d/dx (x)#
#(dp)/(dx) = 4 * 2x - 1#
#(dp)/(dx) = 8x - 1#
Now put it back:
#(dg)/(dx) = (4x^2 - x)(dq)/(dx) + (1/(e^(3x)))(8x - 1)#
Let's now solve for the derivative of #q(x)#:
#(dq)/(dx) = d/dx (1/(e^(3x)))#
We can use the chain rule here. Decompose #q(x)# into several parts:
#q_1(x) = 3x#
#q_2(x) = e^x#
#q_3(x) = 1/x#
So that #q(x) = q_3(q_2(q_1(x)))#. Let's first take the derivative of #q_1(x)#:
#(dq_1)/(dx) = d/dx (3x) = 3#
Also solve for the differential by "multiplying" by #dx#:
#(dq_1)/(dx) = 3 rarr dq_1 = 3 dx#
Now, take the derivative of #q_2(x)# with respect to #q_1(x)#, which means to shove #q_1# as the input:
#(dq_2)/(dq_1) = d/(dq_1) (e^(q_1)) = e^(q_1)#
Solve for the differential #dq_2#:
#(dq_2)/(dq_1) = e^(q_1) rarr dq_2 = e^(q_1) dq_1#
And evaluate #q_1#, as well as #dq_1# as we have solved for earlier:
#dq_2 = e^(3x) * 3dx = 3e^(3x)dx#
Finally, "divide" by #dx# to solve for the derivative:
#(dq_2)/(dx) = 3e^(3x)#
Next, the derivative of #q_3(x)# with respect to #q_2(x)#:
#(dq_3)/(dq_2) = d/(dq_2) (1/x) = - 1/((q_2)^2)#
Solving for the differential:
#(dq_3)/(dq_2) = - 1/((q_2)^2) rarr dq_3 = - 1/((q_2)^2) dq_2#
And evaluating:
#(dq_3)/(dq_2) = - 1/((q_2)^2) rarr dq_3 = - 1/((e^(q_1))^2) * 3e^(3x) dx#
#dq_3 = - (3e^(3x))/((e^(q_1))^2) dx#
Again, for #q_1#:
#dq_3 = - (3e^(3x))/((e^(3x))^2) dx#
Simplifying further:
#dq_3 = - (3)/(e^(3x)) dx#
Finally dividing through:
#(dq_3)/(dx) = -(3)/(e^(3x))#
Since we differentiated all the way, this is the derivative of #q(x)# as a whole:
#(dq)/(dx) = -(3)/(e^(3x))#
Now let's put it back to the derivative of #g(x)#:
#(dg)/(dx) = (4x^2 - x)(dq)/(dx) + (1/(e^(3x)))(8x - 1)#
#(dg)/(dx) = (4x^2 - x)(-(3)/(e^(3x))) + (1/(e^(3x)))(8x - 1)#
And simplify:
#(dg)/(dx) = (4x^2 - x)(-(3)/(e^(3x))) + (8x - 1)/(e^(3x))#
#(dg)/(dx) = -(3(4x^2 - x))/(e^(3x)) + (8x - 1)/(e^(3x))#
#(dg)/(dx) = (3(-4x^2 + x))/(e^(3x)) + (8x - 1)/(e^(3x))#
#(dg)/(dx) = (-12x^2 + 3x)/(e^(3x)) + (8x - 1)/(e^(3x))#
#(dg)/(dx) = ((-12x^2 + 3x) + (8x - 1))/(e^(3x))#
#(dg)/(dx) = (-12x^2 + 11x - 1)/(e^(3x))#
Now that we have the derivative of #g(x)#, let's put it back into the derivative of #f(x)#:
#(df)/(dx) = 42x^2 - (dg)/(dx)#
#(df)/(dx) = 42x^2 - (-12x^2 + 11x - 1)/(e^(3x))#
#(df)/(dx) = 42x^2 + (12x^2 - 11x + 1)/(e^(3x))#
The derivative is the slope of the tangent line for any #x#. Since we have #x = -2#, substitute:
#rarr 42(-2)^2 + (12(-2)^2 - 11(-2) + 1)/(e^(3(-2)))#
#= 42(4) + (12(4) + 22 + 1)/(e^(-6))#
#= 168 + (48 + 23)(e^6)#
#= 168 + 71e^6#
#= 168 + 71e^6#
And that, my friends, is the slope of our tangent line. Hmmh, but if we were to use the slope-intercept form #y = mx + b#, and we know #m = 168 + 71e^6# (for #x = -2#), what is #b#?
Well, we could take a point on the graph (yes, our original function's graph, since this tangent line should touch it) and put in its #x# and #y# coordinates. We could use #x = -2#, but we'd need to solve for #f(x)# to get #y#:
#f(-2) = 14(-2)^3 - (4(-2)^2 - (-2))/(e^(3(-2)))#
#f(-2) = 14(-8) - (4(4) + 2)/(e^(-6))#
#f(-2) = -112 - (16 + 2)(e^6)#
#f(-2) = -112 - 18e^6#
Alright, so #y = -112 - 18e^6#. Let's substitute for what we have:
#y = mx + b rarr -112 - 18e^6 = (168 + 71e^6)(-2) + b#
And solve for #b#:
#b = -112 - 18e^6 + (168 + 71e^6)(2)#
#b = -112 - 18e^6 + 336 + 142e^6#
#b = 224 + 124e^6#
Now that we know #m# and #b#, we could put them back in #y = mx + b# to get:
#y = (168 + 71e^6)x + 224 + 124e^6#
#y = 168x + 71e^6 x + 224 + 124e^6#
Which is the equation of our tangent line at #x = -2#.
