$a+b\sin{x}$ and $b+a\sin{x}$ are two algebraic trigonometric expressions, where $a$ and $b$ are constants, and $x$ represents an angle of a right triangle and also a variable. In this derivative problem, the differentiation of quotient of $a+b\sin{x}$ by $b+a\sin{x}$ has to calculate with respect to $x$.
$= \,\,\,$ $\dfrac{d}{dx}{\, \Bigg(\dfrac{a+b\sin{x}}{b+a\sin{x}}\Bigg)}$
The given function is in quotient form. So, the differentiation of the given function can be calculated by the quotient rule of derivatives. According to Leibniz’s notation, the quotient rule of differentiation can be written in the following form.
$\dfrac{d}{dx}{\, \Big(\dfrac{u}{v}\Big)}$ $\,=\,$ $\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}$
Take $u = a+b\sin{x}$ and $v = b+a\sin{x}$, then start differentiating the given function with respect to $x$.
$= \,\,\,$ $\dfrac{(b+a\sin{x}) \times \dfrac{d}{dx}{\, (a+b\sin{x})}-(a+b\sin{x}) \times \dfrac{d}{dx}{\, (b+a\sin{x})}}{(b+a\sin{x})^2}$
The derivative of sum of two functions can be done by the sum rule of derivatives.
$= \,\,\,$ $\dfrac{(b+a\sin{x}) \times \Big(\dfrac{d}{dx}{\,(a)}+\dfrac{d}{dx}{\,(b\sin{x})\Big)}-(a+b\sin{x}) \times \Big(\dfrac{d}{dx}{\,(b)}+\dfrac{d}{dx}{\,(a\sin{x})\Big)}}{(b+a\sin{x})^2}$
The derivative of a constant is always zero as per the derivative of constant rule.
$= \,\,\,$ $\dfrac{(b+a\sin{x}) \times \Big(0+\dfrac{d}{dx}{\,(b\sin{x})\Big)}-(a+b\sin{x}) \times \Big(0+\dfrac{d}{dx}{\,(a\sin{x})\Big)}}{(b+a\sin{x})^2}$
$= \,\,\,$ $\dfrac{(b+a\sin{x}) \times \dfrac{d}{dx}{\,(b\sin{x})}-(a+b\sin{x}) \times \dfrac{d}{dx}{\,(a\sin{x})}}{(b+a\sin{x})^2}$
In both $b\sin{x}$ and $a\sin{x}$ functions, the factors $a$ and $b$ are constants. So, they can be separated from trigonometric functions by the constant multiple rule of differentiation.
$= \,\,\,$ $\dfrac{(b+a\sin{x}) \times b \times \dfrac{d}{dx}{\,\sin{x}}-(a+b\sin{x}) \times a \times \dfrac{d}{dx}{\,\sin{x}}}{(b+a\sin{x})^2}$
$= \,\,\,$ $\dfrac{b(b+a\sin{x}) \times \dfrac{d}{dx}{\,\sin{x}}-a(a+b\sin{x}) \times \dfrac{d}{dx}{\,\sin{x}}}{(b+a\sin{x})^2}$
According to the derivative of sin function, the derivative of $\sin{x}$ with respect to $x$ is equal to $\cos{x}$.
$= \,\,\,$ $\dfrac{b(b+a\sin{x}) \times \cos{x}-a(a+b\sin{x}) \times \cos{x}}{(b+a\sin{x})^2}$
The differentiation for the given function is successfully completed and now it is time to simplify the function.
$= \,\,\,$ $\dfrac{b(b+a\sin{x})\cos{x}-a(a+b\sin{x})\cos{x}}{(b+a\sin{x})^2}$
In numerator, cosx is a common factor in the both terms of the expression. It can be taken out common from the expression by factorization by taking out common factor.
$= \,\,\,$ $\dfrac{\cos{x}\Big(b(b+a\sin{x})-a(a+b\sin{x})\Big)}{(b+a\sin{x})^2}$
As per distributive property of multiplication over addition, Each constant can be multiplied to its respective factor in the expression of the numerator.
$= \,\,\,$ $\dfrac{\cos{x}\Big(b^2+ab\sin{x}-(a^2+ab\sin{x})\Big)}{(b+a\sin{x})^2}$
Now, simplify the whole function to find the differentiation of the given function mathematically.
$= \,\,\,$ $\dfrac{\cos{x}\Big(b^2+ab\sin{x}-a^2-ab\sin{x}\Big)}{(b+a\sin{x})^2}$
$= \,\,\,$ $\dfrac{\cos{x}\Big(b^2-a^2+ab\sin{x}-ab\sin{x}\Big)}{(b+a\sin{x})^2}$
$= \,\,\,$ $\require{cancel} \dfrac{\cos{x}\Big(b^2-a^2+\cancel{ab\sin{x}}-\cancel{ab\sin{x}}\Big)}{(b+a\sin{x})^2}$
$= \,\,\,$ $\dfrac{\cos{x}(b^2-a^2)}{(b+a\sin{x})^2}$
$= \,\,\,$ $\dfrac{(b^2-a^2)\cos{x}}{(b+a\sin{x})^2}$
Therefore, it is calculated in this calculus problem that the derivative of the ratio of $a+b\sin{x}$ to $b+a\sin{x}$ with respect to $x$ is equal to the quotient of $(b^2-a^2)\cos{x}$ by the square of $b+a\sin{x}$.
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