91Ó°ÊÓ

Understanding the Chain Rule is fundamental in calculus differentiation. The Chain Rule is used when you have a composite function. A composite function is when one function is inside another.
For example, in the exercise, the function was \((z^2 + 2z + 1)^7\). The outer function is \((u)^7\) and the inner function is \(u = z^2 + 2z + 1\).
The Chain Rule states if you have a function \(y = (f(z))^n\), then its derivative will be: \(\frac{d}{dz}\big((f(z))^n\big) = n \big(f(z)\big)^{n-1} \frac{df(z)}{dz}\).
This means you:
1. Take the derivative of the outer function.
2. Multiply it by the derivative of the inner function.
For our function, this became: \(7 (z^2 + 2z + 1)^6 \frac{d}{dz}(z^2 + 2z + 1)\).
This can seem complicated, but always remember: first, identify your inner and outer functions, then apply the rule step by step.

Polynomials are one of the most common functions in calculus, and knowing how to differentiate them is crucial. A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients.
The basic rule for finding the derivative of a polynomial term is to multiply the coefficient by the exponent and then reduce the exponent by one.
For example, if you have \(z^2\), its derivative is \(\frac{d}{dz}(z^2) = 2z\).
Let's look at the inner function from the exercise: \(z^2 + 2z + 1\).
By taking the derivative, we get: \( \frac{d}{dz}(z^2) = 2z \), and \( \frac{d}{dz}(2z) = 2 \), while the derivative of a constant like 1 is zero.
Putting it all together: \( \frac{d}{dz}(z^2 + 2z + 1) = 2z + 2 \).
Combining this with the Chain Rule, we derived: \( 7(z^2 + 2z + 1)^6 (2z + 2) \).

Substitution is a handy technique in calculus, especially when you're asked to find the derivative value at a specific point. After finding the derivative, you'll often need to substitute a specific value into it.
In our exercise, after applying the Chain Rule and differentiating the inner function, we obtained: \(7(z^2+2z+1)^6 (2z+2)\text{\).
Next, we substituted \(z = -1\) into this derivative.
Let's see the steps: \( \frac{d}{dz}\big((z^2 + 2z + 1)^7\big) = 7(z^2 + 2z + 1)^6 (2z + 2)\text{\).
Substituting \(z = -1\): \( 7[(-1)^2 + 2(-1) + 1]^6 (2(-1) + 2)=7[1 - 2 + 1]^6 (2-2)=7[0]^6 \times 0 = 0\text{\).
This shows that the derivative of \((z^2 + 2z + 1)^7\) at \(z = -1\text{ is 0.\text{\).
The substitution step is crucial as it gives you the exact value of the derivative at a certain point, simplifying your expression and finding your answer.}