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Question: If \[y=\dfrac{1}{\left( a-z \right)}\], then \[\dfrac{dz}{dy}\] is? 1.\[{{\left( a-z \right)}^{2}}...

If y=1(a−z)y=\dfrac{1}{\left( a-z \right)}, then dzdy\dfrac{dz}{dy} is?
1.(a−z)2{{\left( a-z \right)}^{2}}
2.−(z−a)2-{{\left( z-a \right)}^{2}}
3.(z+a)2{{\left( z+a \right)}^{2}}
4.−(z+a)2-{{\left( z+a \right)}^{2}}

Explanation

Solution

Since we are supposed to differentiate y=1(a−z)y=\dfrac{1}{\left( a-z \right)}, with respect to yy, firstly, we will be transposing the term (a−z)\left( a-z \right) to the left hand side of the equation and solve it there. Then we will be transposing all the terms except zz right back to the right hand side. Then we will be differentiating the obtained equation with respect to yy and that would be our required solution.

Complete step by step answer:
Now let us learn more about differentiation. It is one of the important parts of calculus. Differentiation is nothing but finding out the instantaneous rate of change of function based on any one of its variables. The opposite of differentiation is anti-differentiation. There are different rules in differentiation to be followed while computing. They are: Sum and Difference rule, Multiplication rule, Division rule and Chain rule. while differentiating we generally follow Leibniz's notation and i.e. dydx\dfrac{dy}{dx}.
Now let start solving the given problem;
The function given is y=1(a−z)y=\dfrac{1}{\left( a-z \right)}.
Firstly, let us cross multiply the terms. Upon doing so, we get

& y=\dfrac{1}{\left( a-z \right)} \\\ & \Rightarrow y\left( a-z \right)=1 \\\ & \Rightarrow ay-zy=1 \\\ \end{aligned}$$ Now we will be transposing the $$ay$$ term to the RHS. $$\begin{aligned} & \Rightarrow ay-zy=1 \\\ & \Rightarrow ay-1=zy \\\ \end{aligned}$$ Upon further solving it so as to differentiate it with respect to $$y$$, we get $$\begin{aligned} & \Rightarrow ay-1=zy \\\ & \Rightarrow z=\dfrac{ay-1}{y}=a-\dfrac{1}{y} \\\ \end{aligned}$$ Now we will be differentiating with respect to $$y$$. $$\Rightarrow \dfrac{dz}{dy}=0+\dfrac{1}{{{y}^{2}}}$$ (by applying the reciprocal rule) Since we know that $$y=\dfrac{1}{\left( a-z \right)}$$, upon substituting we get, $$\Rightarrow {{\left( a-z \right)}^{2}}$$ **So, the correct answer is “Option 1”.** **Note:** While differentiating, we must check for the variable for which the function is to be differentiated. Not checking it is the common mistake committed. We must have a grip of the rules to be applied appropriately. Differentiation can be applied in finding out the acceleration, to find tangent to a curve etc.