Math Doubts

Find $\dfrac{d}{dx}{\, \Big(\dfrac{a+b\sin{x}}{b+a\sin{x}}\Big)}$

$a+b\sin{x}$ 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 $x$.

$= \,\,\,$ $\dfrac{d}{dx}{\, \Bigg(\dfrac{a+b\sin{x}}{b+a\sin{x}}\Bigg)}$

Use quotient rule of derivatives

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 $x$.

$= \,\,\,$ $\dfrac{(b+a\sin{x}) \times \dfrac{d}{dx}{\, (a+b\sin{x})}-(a+b\sin{x}) \times \dfrac{d}{dx}{\, (b+a\sin{x})}}{(b+a\sin{x})^2}$

Start the process of the Differentiation

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 $\cos{x}$.

$= \,\,\,$ $\dfrac{b(b+a\sin{x}) \times \cos{x}-a(a+b\sin{x}) \times \cos{x}}{(b+a\sin{x})^2}$

Simplify the Algebraic trigonometric function

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 $b+a\sin{x}$.

Math Doubts
Math Doubts is a free math tutor for helping students to learn mathematics online from basics to advanced scientific level for teachers to improve their teaching skill and for researchers to share their research projects. Know more
Follow us on Social Media
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more