Find $\frac{\mathrm{dz}}{\mathrm{dt}}$ given $\mathrm{z}=x \mathrm{y}^{2}$ and $\mathrm{x}=(\mathrm{t}-1)$ and $y=\frac{1}{(t-1)^{2}}$ .

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Algebra 2

Evaluate expression when two complex numbers are given

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Q. Find

\frac{\mathrm{dz}}{\mathrm{dt}}

given

\mathrm{z}=x \mathrm{y}^{2}

and

\mathrm{x}=(\mathrm{t}-1)

and

y=\frac{1}{(t-1)^{2}}

Find $\frac{dz}{dt}$ using product rule: First, let's find $\frac{dz}{dt}$ using the product rule for derivatives, which is $\frac{d(uv)}{dt} = u\left(\frac{dv}{dt}\right) + v\left(\frac{du}{dt}\right)$ , where $z=xy^2$ , $u=x$ , and $v=y^2$ .
Differentiate $x$ and $y^2$ with respect to $t$ : Differentiate $x$ with respect to $t$ , $\frac{dx}{dt} = \frac{d(t-1)}{dt} = 1$ .
Find $\frac{dy}{dt}$ : Differentiate $y^2$ with respect to $t$ , using the chain rule, $\frac{d(y^2)}{dt} = 2y \cdot \frac{dy}{dt}$ .
Substitute $\frac{dy}{dt}$ into derivative of $y^2$ : Now, find $\frac{dy}{dt}$ . $y = \frac{1}{(t-1)^2}$ , so $\frac{dy}{dt} = \frac{d(\frac{1}{(t-1)^2})}{dt}$ .
Simplify $\frac{d y^2}{d t}$ : Using the chain rule and the power rule, $\frac{d y}{d t} = -2 \times \frac{1}{(t-1)^3} \times \frac{d(t-1)}{d t} = -\frac{2}{(t-1)^3}$ .
Apply product rule to find $\frac{dz}{dt}$ : Substitute $\frac{dy}{dt}$ back into the derivative of $y^2$ , $\frac{d(y^2)}{dt} = 2y \cdot \frac{dy}{dt} = 2 \cdot \frac{1}{(t-1)^2} \cdot -\frac{2}{(t-1)^3}$ .
Substitute values into $\frac{dz}{dt}$ equation: Simplify $\frac{d(y^2)}{dt}$ , $\frac{d(y^2)}{dt} = \frac{-4}{(t-1)^5}$ .
Simplify $\frac{dz}{dt}$ : Now, apply the product rule to find $\frac{dz}{dt}$ , $\frac{dz}{dt} = x\left(\frac{d(y^2)}{dt}\right) + y^2\left(\frac{dx}{dt}\right)$ .
Combine like terms: Substitute $x$ , $y^2$ , rac{dx}{dt}, and rac{dy^2}{dt} into the equation, $rac{dz}{dt} = (t$-1$)igg(-rac{$4$}{(t$-1$)^$5$}igg) + igg(rac{$1$}{(t$-1$)^$4$}igg)($1$).
Finish simplifying $\frac{dz}{dt}$: Simplify $\frac{dz}{dt}$, $\frac{dz}{dt} = \frac{-4}{(t-1)^4} + \frac{1}{(t-1)^4}$.
Finish simplifying $\frac{dz}{dt}$: Simplify $\frac{dz}{dt}$, $\frac{dz}{dt} = \frac{-4}{(t-1)^4} + \frac{1}{(t-1)^4}$.Combine like terms, $\frac{dz}{dt} = \frac{(-4 + 1)}{(t-1)^4}$.
Finish simplifying $\frac{dz}{dt}$: Simplify $\frac{dz}{dt}$, $\frac{dz}{dt} = -\frac{4}{(t-1)^4} + \frac{1}{(t-1)^4}$.Combine like terms, $\frac{dz}{dt} = \frac{-4 + 1}{(t-1)^4}$.Finish simplifying, $\frac{dz}{dt} = -\frac{3}{(t-1)^4}$.