\(y = tan^{-1} (\frac{4 sin 2x}{cos 2x - 6sin^2x})\) then \(\frac{dy}{dx}\) at \(x=0\).
Solution and Explanation
To find \(\frac{dx}{dt}\) of \(tan^{-1}[\frac{(cos(2x)-6sin^2(x))}{(4sin^2(x))}]\) at \(x=0\),
we first substitute \(u=\frac{(cos(2x)-6sin^2(x))}{(4sin^2(x))}\).
Then, using the chain rule, we get \(\frac{dx}{dt}\) of \(tan^{-1}(u)\) times \(\frac{dx}{dt}\) of u.
Simplifying the expression and evaluating it at \(x=0\), we get the final answer as 8.
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
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