# Evaluate $\dfrac{d}{dx}{\Bigg[\dfrac{1}{b}\tan^{-1}{\Big(\dfrac{x}{b}\Big)}-\dfrac{1}{a}\tan^{-1}{\Big(\dfrac{x}{a}\Big)}\Bigg]}$

An inverse trigonometric expression is formed by the inverse tangent functions. We have to evaluate the differentiation of the inverse trigonometric expression with respect to $x$.

$\dfrac{d}{dx}{\Bigg[\dfrac{1}{b}\,\tan^{-1}{\Big(\dfrac{x}{b}\Big)}-\dfrac{1}{a}\,\tan^{-1}{\Big(\dfrac{x}{a}\Big)}\Bigg]}$

### Difference Rule of Derivatives

The derivative of difference of two terms can be evaluated by the difference of their derivatives as per the difference rule of derivatives.

$= \,\,\,$ $\dfrac{d}{dx}{\Bigg(\dfrac{1}{b}\,\tan^{-1}{\Big(\dfrac{x}{b}\Big)}\Bigg)}$ $-$ $\dfrac{d}{dx}{\Bigg(\dfrac{1}{a}\,\tan^{-1}{\Big(\dfrac{x}{a}\Big)}}\Bigg)$

### Separate Constants from Differentiation

In this differentiation problem, $x$ is a variable and the remaining literals are constants. Hence, the factors, which are constants can be separated from the differentiation by the constant multiple rule of differentiation.

$= \,\,\,$ $\dfrac{1}{b} \times \dfrac{d}{dx}{\,\tan^{-1}{\Big(\dfrac{x}{b}\Big)}}$ $-$ $\dfrac{1}{a} \times \dfrac{d}{dx}{\,\tan^{-1}{\Big(\dfrac{x}{a}\Big)}}$

### Evaluate the Differentiation of Each term

The differentiation of each inverse tan function in the expression can be evaluated in two ways.

#### Beginner’s Method

Let $y \,=\, \dfrac{x}{b}$ and $z \,=\, \dfrac{x}{a}$, then differentiate each equation with respect to $x$.

$\implies$ $\dfrac{dy}{dx} \,=\, \dfrac{d}{dx}{\,\Big(\dfrac{x}{b}\Big)}$

$\implies$ $\dfrac{dy}{dx} \,=\, \dfrac{d}{dx}{\,\Big(\dfrac{1}{b} \times x\Big)}$

$\implies$ $\dfrac{dy}{dx} \,=\, \dfrac{1}{b} \times \dfrac{d}{dx}{\,(x)}$

$\implies$ $\dfrac{dy}{dx} \,=\, \dfrac{1}{b} \times \dfrac{dx}{dx}$

$\implies$ $\dfrac{dy}{dx} \,=\, \dfrac{1}{b} \times 1$

$\implies$ $\dfrac{dy}{dx} \,=\, \dfrac{1}{b}$

$\implies$ $b \times dy \,=\, 1 \times dx$

$\implies$ $bdy \,=\, dx$

$\,\,\,\therefore \,\,\,\,\,\,$ $dx \,=\, bdy$

Similarly, we can differentiate the second equation as follows.

$\implies$ $\dfrac{dz}{dx} \,=\, \dfrac{d}{dx}{\,\Big(\dfrac{x}{a}\Big)}$

$\implies$ $\dfrac{dz}{dx} \,=\, \dfrac{d}{dx}{\,\Big(\dfrac{1}{a} \times x\Big)}$

$\implies$ $\dfrac{dz}{dx} \,=\, \dfrac{1}{a} \times \dfrac{d}{dx}{\,(x)}$

$\implies$ $\dfrac{dz}{dx} \,=\, \dfrac{1}{a} \times \dfrac{dx}{dx}$

$\implies$ $\dfrac{dz}{dx} \,=\, \dfrac{1}{a} \times 1$

$\implies$ $\dfrac{dz}{dx} \,=\, \dfrac{1}{a}$

$\implies$ $a \times dz \,=\, 1 \times dx$

$\implies$ $adz \,=\, dx$

$\,\,\,\therefore \,\,\,\,\,\,$ $dx \,=\, adz$

Now, comeback to the differential expression in inverse trigonometric functions form.

$= \,\,\,$ $\dfrac{1}{b} \times \dfrac{d}{dx}{\,\tan^{-1}{\Big(\dfrac{x}{b}\Big)}}$ $-$ $\dfrac{1}{a} \times \dfrac{d}{dx}{\,\tan^{-1}{\Big(\dfrac{x}{a}\Big)}}$

We have taken that $y \,=\, \dfrac{x}{b}$ and $z \,=\, \dfrac{x}{a}$, and also obtained that $dx \,=\, bdy$ and $dx \,=\, adz$ by differentiating both equations. Now, express the differential expression in terms of $y$ and $z$.

$= \,\,\,$ $\dfrac{1}{b} \times \dfrac{d}{bdy}{\,\tan^{-1}{\,(y)}}$ $-$ $\dfrac{1}{a} \times \dfrac{d}{adz}{\,\tan^{-1}{\,(z)}}$

$= \,\,\,$ $\dfrac{1}{b} \times \dfrac{d}{b \times dy}{\,\tan^{-1}{\,(y)}}$ $-$ $\dfrac{1}{a} \times \dfrac{d}{a \times dz}{\,\tan^{-1}{\,(z)}}$

$= \,\,\,$ $\dfrac{1}{b} \times \dfrac{1}{b} \times \dfrac{d}{dy}{\,\tan^{-1}{\,(y)}}$ $-$ $\dfrac{1}{a} \times \dfrac{1}{a} \times \dfrac{d}{dz}{\,\tan^{-1}{\,(z)}}$

$= \,\,\,$ $\dfrac{1 \times 1}{b \times b} \times \dfrac{d}{dy}{\,\tan^{-1}{\,(y)}}$ $-$ $\dfrac{1 \times 1}{a \times a} \times \dfrac{d}{dz}{\,\tan^{-1}{\,(z)}}$

$= \,\,\,$ $\dfrac{1}{b^2} \times \dfrac{d}{dy}{\,\tan^{-1}{\,(y)}}$ $-$ $\dfrac{1}{a^2} \times \dfrac{d}{dz}{\,\tan^{-1}{\,(z)}}$

Now, evaluate differentiation of each term by the derivative rule of inverse tan function.

$= \,\,\,$ $\dfrac{1}{b^2} \times \dfrac{1}{1+y^2}$ $-$ $\dfrac{1}{a^2} \times \dfrac{1}{1+z^2}$

Now, replace the values of $y$ and $z$ to get the solution in $x$.

$= \,\,\,$ $\dfrac{1}{b^2} \times \dfrac{1}{1+\Big(\dfrac{x}{b}\Big)^2}$ $-$ $\dfrac{1}{a^2} \times \dfrac{1}{1+\Big(\dfrac{x}{a}\Big)^2}$

Now, simplify the algebraic expression to get the required result of the given derivative problem.

$= \,\,\,$ $\dfrac{1}{b^2} \times \dfrac{1}{1+\dfrac{x^2}{b^2}}$ $-$ $\dfrac{1}{a^2} \times \dfrac{1}{1+\dfrac{x^2}{a^2}}$

$= \,\,\,$ $\dfrac{1}{b^2} \times \dfrac{1}{\dfrac{1 \times b^2+x^2}{b^2}}$ $-$ $\dfrac{1}{a^2} \times \dfrac{1}{\dfrac{1\times a^2+x^2}{a^2}}$

$= \,\,\,$ $\dfrac{1}{b^2} \times \dfrac{1}{\dfrac{b^2+x^2}{b^2}}$ $-$ $\dfrac{1}{a^2} \times \dfrac{1}{\dfrac{a^2+x^2}{a^2}}$

$= \,\,\,$ $\dfrac{1}{b^2} \times \dfrac{1}{\dfrac{x^2+b^2}{b^2}}$ $-$ $\dfrac{1}{a^2} \times \dfrac{1}{\dfrac{x^2+a^2}{a^2}}$

$= \,\,\,$ $\dfrac{1}{b^2} \times \dfrac{b^2}{x^2+b^2}$ $-$ $\dfrac{1}{a^2} \times \dfrac{a^2}{x^2+a^2}$

$= \,\,\,$ $\dfrac{1 \times b^2}{b^2 \times (x^2+b^2)}$ $-$ $\dfrac{1 \times a^2}{a^2 \times (x^2+a^2)}$

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

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{b^2}}{\cancel{b^2}(x^2+b^2)}$ $-$ $\dfrac{\cancel{a^2}}{\cancel{a^2}(x^2+a^2)}$

$= \,\,\,$ $\dfrac{1}{x^2+b^2}$ $-$ $\dfrac{1}{x^2+a^2}$

$= \,\,\,$ $\dfrac{1 \times (x^2+a^2)-1 \times (x^2+b^2)}{(x^2+b^2)(x^2+a^2)}$

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

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

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

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

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

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

The following differential expression can also be differentiated in an advanced method.

$= \,\,\,$ $\dfrac{1}{b} \times \dfrac{d}{dx}{\,\tan^{-1}{\Big(\dfrac{x}{b}\Big)}}$ $-$ $\dfrac{1}{a} \times \dfrac{d}{dx}{\,\tan^{-1}{\Big(\dfrac{x}{a}\Big)}}$

Each term can be differentiated by the chain rule of differentiation.

$= \,\,\,$ $\dfrac{1}{b} \times \dfrac{1}{1+\Big(\dfrac{x}{b}\Big)^2} \times \dfrac{d}{dx}{\,\Big(\dfrac{x}{b}\Big)}$ $-$ $\dfrac{1}{a} \times \dfrac{1}{1+\Big(\dfrac{x}{a}\Big)^2} \times \dfrac{d}{dx}{\,\Big(\dfrac{x}{a}\Big)}$

$= \,\,\,$ $\dfrac{1}{b} \times \dfrac{1}{1+\Big(\dfrac{x}{b}\Big)^2} \times \dfrac{d}{dx}{\,\Big(\dfrac{1}{b} \times x\Big)}$ $-$ $\dfrac{1}{a} \times \dfrac{1}{1+\Big(\dfrac{x}{a}\Big)^2} \times \dfrac{d}{dx}{\,\Big(\dfrac{1}{a} \times x\Big)}$

$= \,\,\,$ $\dfrac{1}{b} \times \dfrac{1}{1+\Big(\dfrac{x}{b}\Big)^2} \times \dfrac{1}{b} \times \dfrac{d}{dx}{\,(x)}$ $-$ $\dfrac{1}{a} \times \dfrac{1}{1+\Big(\dfrac{x}{a}\Big)^2} \times \dfrac{1}{a} \times \dfrac{d}{dx}{\,(x)}$

Now, evaluate the derivative of variable with respect to same variable.

$= \,\,\,$ $\dfrac{1}{b} \times \dfrac{1}{1+\Big(\dfrac{x}{b}\Big)^2} \times \dfrac{1}{b} \times 1$ $-$ $\dfrac{1}{a} \times \dfrac{1}{1+\Big(\dfrac{x}{a}\Big)^2} \times \dfrac{1}{a} \times 1$

Finally, simplify the algebraic expression to obtain the required result of the differentiation of this problem.

$= \,\,\,$ $\dfrac{1}{b} \times \dfrac{1}{1+\Big(\dfrac{x}{b}\Big)^2} \times \dfrac{1}{b}$ $-$ $\dfrac{1}{a} \times \dfrac{1}{1+\Big(\dfrac{x}{a}\Big)^2} \times \dfrac{1}{a}$

$= \,\,\,$ $\dfrac{1}{b} \times \dfrac{1}{b} \times \dfrac{1}{1+\Big(\dfrac{x}{b}\Big)^2}$ $-$ $\dfrac{1}{a} \times \dfrac{1}{a} \times \dfrac{1}{1+\Big(\dfrac{x}{a}\Big)^2}$

$= \,\,\,$ $\dfrac{1 \times 1}{b \times b}\times \dfrac{1}{1+\Big(\dfrac{x}{b}\Big)^2}$ $-$ $\dfrac{1 \times 1}{a \times a} \times \dfrac{1}{1+\Big(\dfrac{x}{a}\Big)^2}$

$= \,\,\,$ $\dfrac{1}{b^2}\times \dfrac{1}{1+\Big(\dfrac{x}{b}\Big)^2}$ $-$ $\dfrac{1}{a^2} \times \dfrac{1}{1+\Big(\dfrac{x}{a}\Big)^2}$

$= \,\,\,$ $\dfrac{1}{b^2}\times \dfrac{1}{1+\dfrac{x^2}{b^2}}$ $-$ $\dfrac{1}{a^2} \times \dfrac{1}{1+\dfrac{x^2}{a^2}}$

$= \,\,\,$ $\dfrac{1}{b^2}\times \dfrac{1}{\dfrac{b^2+x^2}{b^2}}$ $-$ $\dfrac{1}{a^2} \times \dfrac{1}{\dfrac{a^2+x^2}{a^2}}$

$= \,\,\,$ $\dfrac{1}{b^2}\times \dfrac{1}{\dfrac{x^2+b^2}{b^2}}$ $-$ $\dfrac{1}{a^2} \times \dfrac{1}{\dfrac{x^2+a^2}{a^2}}$

$= \,\,\,$ $\dfrac{1}{b^2}\times \dfrac{b^2}{x^2+b^2}$ $-$ $\dfrac{1}{a^2} \times \dfrac{a^2}{x^2+a^2}$

$= \,\,\,$ $\dfrac{1 \times b^2}{b^2 \times (x^2+b^2)}$ $-$ $\dfrac{1 \times a^2}{a^2 \times (x^2+a^2)}$

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

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{b^2}}{\cancel{b^2}(x^2+b^2)}$ $-$ $\dfrac{\cancel{a^2}}{\cancel{a^2}(x^2+a^2)}$

$= \,\,\,$ $\dfrac{1}{x^2+b^2}$ $-$ $\dfrac{1}{x^2+a^2}$

$= \,\,\,$ $\dfrac{1 \times (x^2+a^2)-1 \times (x^2+b^2)}{(x^2+b^2)(x^2+a^2)}$

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

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

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

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

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

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

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