Analyzing the Setup
Imagine standing before a massive, intimidating mountain. If you try to climb it straight up the sheer face, you will likely fall. But if you look for the winding path, the climb becomes a journey.
We are given the function:
y=loge(1+x21−x2)
If you jump straight into the quotient rule, you are climbing the sheer face. Instead, let's use the power of logarithmic properties. We know that loge(ba)=loge(a)−loge(b).
This simple identity transforms our function into:
y=loge(1−x2)−loge(1+x2)
Now, the mountain is just a gentle hill.
The First Derivative
Now that we have simplified, let's find y′. We apply the chain rule to each term.
The derivative of
loge(1−x2) is
1−x2−2x, and the derivative of
loge(1+x2) is
1+x22x. When we combine these, we get:
y′=1−x2−2x−1+x22x
Factoring out −2x, we find a common denominator of (1−x2)(1+x2), which is 1−x4. The numerator simplifies beautifully to (1+x2)+(1−x2)=2.
Thus, the first derivative is:
y′=1−x4−4x
The Second Derivative
Now, we face the second derivative, y′′. We must differentiate 1−x4−4x using the quotient rule: v2u′v−uv′.
Here,
u=−4x and
v=1−x4. The derivatives are
u′=−4 and
v′=−4x3. Plugging these in, we get:
y′′=(1−x4)2−4(1−x4)−(−4x)(−4x3)
Simplifying the numerator, we distribute the −4 to get −4+4x4, and the second term becomes −16x4. Combining these, we get −4−12x4, or −4(1+3x4).
So, the second derivative is:
y′′=(1−x4)2−4(1+3x4)
The Final Evaluation
We are at the finish line. We need to evaluate 225(y′−y′′) at x=21.
First, let's find the value of x4=(21)4=161. Then 1−x4=1−161=1615.
Now, substitute these into our derivatives:
For
y′, we get:
y′=15/16−4(1/2)=15/16−2=−1532
For
y′′, we get:
y′′=(15/16)2−4(1+3/16)=225/256−4(19/16)=225/256−19/4=−2251216
Now, compute
y′−y′′:
y′−y′′=−1532−(−2251216)
Converting to a common denominator of
225, we get:
225−480+1216=225736
Finally, multiply by
225:
225×225736=736
The cancellation is perfect. The final answer is 736.