section 2.5: problem 7 (1 point) compute the derivative of the given function in two different ways…

Explanation:

Step1: Identify f(t) and g(t) for Quotient Rule

Let $f(t)=t^{2}-1$ and $g(t)=t - 1$. First, find $f'(t)$ and $g'(t)$. The derivative of $f(t)=t^{2}-1$ is $f'(t) = 2t$ using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, and the derivative of $g(t)=t - 1$ is $g'(t)=1$.

Step2: Apply the Quotient Rule

The Quotient Rule is $\left[\frac{f(t)}{g(t)}\right]'=\frac{g(t)\cdot f'(t)-f(t)\cdot g'(t)}{[g(t)]^{2}}$. Substitute $f(t), f'(t), g(t), g'(t)$ into the formula: [ \begin{align*} f'(t)&=\frac{(t - 1)\cdot(2t)-(t^{2}-1)\cdot(1)}{(t - 1)^{2}}\ &=\frac{2t^{2}-2t - t^{2}+1}{(t - 1)^{2}}\ &=\frac{t^{2}-2t + 1}{(t - 1)^{2}}\ &=\frac{(t - 1)^{2}}{(t - 1)^{2}}\ &=1 \end{align*} ]

Step3: Simplify f(t) first

We can simplify $f(t)=\frac{t^{2}-1}{t - 1}$. Since $t^{2}-1=(t + 1)(t - 1)$ (using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$), then $f(t)=\frac{(t + 1)(t - 1)}{t - 1}=t + 1$ for $t\neq1$.

Step4: Differentiate the simplified f(t)

The derivative of $f(t)=t + 1$ using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$ (where for $t^1$, $n = 1$ and for the constant 1, the derivative is 0) is $f'(t)=1$.

Answer:

a) $f'(t)=\frac{(t - 1)\cdot(2t)-(t^{2}-1)\cdot(1)}{(t - 1)^{2}}=1$ b) $f'(t)=1$