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Question: If $y = \tan^{-1}(\frac{2+3x}{3-2x}) + \tan^{-1}(\frac{4x}{1+5x^2})$, then $\frac{dy}{dx} =$...

If y=tan⁡−1(2+3x3−2x)+tan⁡−1(4x1+5x2)y = \tan^{-1}(\frac{2+3x}{3-2x}) + \tan^{-1}(\frac{4x}{1+5x^2}), then dydx=\frac{dy}{dx} =

Answer

\frac{5}{1+25x^2}

Explanation

Solution

Solution:

We are given

y=tan⁡−1(2+3x3−2x)+tan⁡−1(4x1+5x2).y=\tan^{-1}\left(\frac{2+3x}{3-2x}\right)+\tan^{-1}\left(\frac{4x}{1+5x^2}\right).

Step 1. Differentiate the first term

Let

A(x)=2+3x3−2x.A(x)=\frac{2+3x}{3-2x}.

Using the quotient rule:

A′(x)=(3)(3−2x)−(2+3x)(−2)(3−2x)2=9−6x+4+6x(3−2x)2=13(3−2x)2.A'(x)=\frac{(3)(3-2x) - (2+3x)(-2)}{(3-2x)^2} =\frac{9-6x+4+6x}{(3-2x)^2}=\frac{13}{(3-2x)^2}.

Then, using the derivative of tan⁡−1u=u′1+u2\tan^{-1} u = \frac{u'}{1+u^2}:

ddxtan⁡−1(A(x))=A′(x)1+A(x)2=13/(3−2x)21+(2+3x3−2x)2.\frac{d}{dx}\tan^{-1}(A(x)) = \frac{A'(x)}{1+A(x)^2} =\frac{13/(3-2x)^2}{1+\left(\frac{2+3x}{3-2x}\right)^2}.

Notice that

1+(2+3x3−2x)2=(3−2x)2+(2+3x)2(3−2x)2.1+\left(\frac{2+3x}{3-2x}\right)^2 =\frac{(3-2x)^2+(2+3x)^2}{(3-2x)^2}.

Thus,

ddxtan⁡−1(A(x))=13(3−2x)2+(2+3x)2.\frac{d}{dx}\tan^{-1}(A(x)) = \frac{13}{(3-2x)^2+(2+3x)^2}.

Compute the denominator:

(3−2x)2+(2+3x)2=(9−12x+4x2)+(4+12x+9x2)=13+13x2=13(1+x2).(3-2x)^2+(2+3x)^2 = (9-12x+4x^2)+(4+12x+9x^2) = 13+13x^2=13(1+x^2).

So,

ddxtan⁡−1(A(x))=1313(1+x2)=11+x2.\frac{d}{dx}\tan^{-1}(A(x)) = \frac{13}{13(1+x^2)} =\frac{1}{1+x^2}.

Step 2. Differentiate the second term

Let

B(x)=4x1+5x2.B(x)=\frac{4x}{1+5x^2}.

Using the quotient rule:

B′(x)=4(1+5x2)−4x(10x)(1+5x2)2=4+20x2−40x2(1+5x2)2=4(1−5x2)(1+5x2)2.B'(x)=\frac{4(1+5x^2)-4x(10x)}{(1+5x^2)^2} =\frac{4+20x^2-40x^2}{(1+5x^2)^2} =\frac{4(1-5x^2)}{(1+5x^2)^2}.

Then,

ddxtan⁡−1(B(x))=B′(x)1+B(x)2.\frac{d}{dx}\tan^{-1}(B(x)) = \frac{B'(x)}{1+B(x)^2}.

Since

B(x)2=16x2(1+5x2)2,B(x)^2=\frac{16x^2}{(1+5x^2)^2},

we have

1+B(x)2=(1+5x2)2+16x2(1+5x2)2.1+B(x)^2=\frac{(1+5x^2)^2+16x^2}{(1+5x^2)^2}.

Thus,

ddxtan⁡−1(B(x))=4(1−5x2)/(1+5x2)2[(1+5x2)2+16x2]/(1+5x2)2=4(1−5x2)(1+5x2)2+16x2.\frac{d}{dx}\tan^{-1}(B(x)) = \frac{4(1-5x^2)/(1+5x^2)^2}{\big[(1+5x^2)^2+16x^2\big]/(1+5x^2)^2} =\frac{4(1-5x^2)}{(1+5x^2)^2+16x^2}.

Simplify the denominator:

(1+5x2)2+16x2=1+10x2+25x4+16x2=1+26x2+25x4.(1+5x^2)^2+16x^2 = 1+10x^2+25x^4+16x^2 = 1+26x^2+25x^4.

Notice that

1+26x2+25x4=(1+x2)(1+25x2).1+26x^2+25x^4 = (1+x^2)(1+25x^2).

Thus,

ddxtan⁡−1(B(x))=4(1−5x2)(1+x2)(1+25x2).\frac{d}{dx}\tan^{-1}(B(x)) = \frac{4(1-5x^2)}{(1+x^2)(1+25x^2)}.

Step 3. Combine the derivatives

The total derivative is:

dydx=11+x2+4(1−5x2)(1+x2)(1+25x2)=1+4(1−5x2)1+25x21+x2.\frac{dy}{dx}=\frac{1}{1+x^2}+\frac{4(1-5x^2)}{(1+x^2)(1+25x^2)} =\frac{1+ \frac{4(1-5x^2)}{1+25x^2}}{1+x^2}.

Writing the numerator with a common denominator:

=(1+25x2)+4(1−5x2)1+25x21+x2=(1+25x2)+4(1−5x2)(1+x2)(1+25x2).= \frac{\frac{(1+25x^2)+4(1-5x^2)}{1+25x^2}}{1+x^2} =\frac{(1+25x^2)+4(1-5x^2)}{(1+x^2)(1+25x^2)}.

Simplify the numerator:

(1+25x2)+4(1−5x2)=1+25x2+4−20x2=5+5x2=5(1+x2).(1+25x^2)+4(1-5x^2)=1+25x^2+4-20x^2=5+5x^2=5(1+x^2).

Cancel (1+x2)(1+x^2):

dydx=51+25x2.\frac{dy}{dx}=\frac{5}{1+25x^2}.

Answer:

dydx=51+25x2\boxed{\frac{dy}{dx}=\frac{5}{1+25x^2}}