GSEB Class 12 Maths Solutions Chapter 5 Continuity and Differentiability Exercise 5.3

Question 1. 2x + 3y = sin x
Answer: We begin by differentiating the given equation with respect to \(x\).
\(2x + 3y = \sin x\)
Differentiating both sides with respect to \(x\):
\( \frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin x) \)
\( 2 + 3\frac{dy}{dx} = \cos x \)
Now, we rearrange the terms to find \( \frac{dy}{dx} \):
\( 3\frac{dy}{dx} = \cos x - 2 \)
\( \frac{dy}{dx} = \frac{\cos x - 2}{3} \)
Alternatively, this can be written as:
\( \frac{dy}{dx} = \frac{1}{3}(\cos x - 2) \)
In simple words: We find how quickly \(y\) changes when \(x\) changes by using differentiation. We move all terms to isolate \( \frac{dy}{dx} \).

Exam Tip: Remember to apply the chain rule when differentiating terms involving \(y\) with respect to \(x\), such as \( \frac{d}{dx}(f(y)) = f'(y) \cdot \frac{dy}{dx} \).

Question 2. 2x + 3y = sin y
Answer: To find the derivative, we differentiate both sides of the equation with respect to \(x\).
\(2x + 3y = \sin y\)
Differentiating both sides with respect to \(x\):
\( \frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y) \)
\( 2 + 3\frac{dy}{dx} = \cos y \cdot \frac{dy}{dx} \)
Next, we gather all terms containing \( \frac{dy}{dx} \) on one side:
\( 2 = \cos y \frac{dy}{dx} - 3\frac{dy}{dx} \)
\( 2 = (\cos y - 3)\frac{dy}{dx} \)
Finally, we solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{2}{\cos y - 3} \)
In simple words: We differentiate both sides, collecting the \( \frac{dy}{dx} \) terms to one side, and then solve to find the expression for \( \frac{dy}{dx} \).

Exam Tip: When differentiating an implicit function like this, remember to apply the chain rule to all \(y\) terms, multiplying by \( \frac{dy}{dx} \).

Question 3. ax + by² = cos y
Answer: We will differentiate the given equation implicitly with respect to \(x\).
\(ax + by^2 = \cos y\)
Differentiating both sides with respect to \(x\):
\( \frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y) \)
\( a + 2by\frac{dy}{dx} = -\sin y \frac{dy}{dx} \)
Now, we rearrange the equation to isolate \( \frac{dy}{dx} \):
\( a = -\sin y \frac{dy}{dx} - 2by\frac{dy}{dx} \)
\( a = -(2by + \sin y)\frac{dy}{dx} \)
To get \( \frac{dy}{dx} \) by itself, we divide:
\( \frac{dy}{dx} = \frac{-a}{2by + \sin y} \)
In simple words: We find the derivative by taking the derivative of each part. We then move all parts with \( \frac{dy}{dx} \) to one side to solve for it.

Exam Tip: Be careful with signs, especially when moving terms across the equals sign or when differentiating trigonometric functions like \( \cos y \).

Question 4. xy + y² = tan x + y
Answer: We need to differentiate the given implicit function with respect to \(x\).
\(xy + y^2 = \tan x + y\)
Differentiating both sides with respect to \(x\), applying the product rule for \(xy\):
\( \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\tan x) + \frac{d}{dx}(y) \)
\( (1 \cdot y + x \frac{dy}{dx}) + 2y\frac{dy}{dx} = \sec^2 x + \frac{dy}{dx} \)
Now, we collect all the terms containing \( \frac{dy}{dx} \) on one side:
\( x\frac{dy}{dx} + 2y\frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x - y \)
Factor out \( \frac{dy}{dx} \):
\( (x + 2y - 1)\frac{dy}{dx} = \sec^2 x - y \)
Finally, solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1} \)
In simple words: We differentiate each term, remembering the product rule for \(xy\) and chain rule for \(y\). Then we group \( \frac{dy}{dx} \) terms and solve.

Exam Tip: The product rule \( \frac{d}{dx}(uv) = u'v + uv' \) is essential for terms like \(xy\). Also, remember to factor out \( \frac{dy}{dx} \) carefully.

Question 5. x² + xy + y² = 100
Answer: We differentiate the equation implicitly with respect to \(x\).
\(x^2 + xy + y^2 = 100\)
Differentiating both sides with respect to \(x\):
\( \frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(100) \)
\( 2x + (1 \cdot y + x \frac{dy}{dx}) + 2y \frac{dy}{dx} = 0 \)
\( 2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \)
Group the terms with \( \frac{dy}{dx} \):
\( (x + 2y)\frac{dy}{dx} = -2x - y \)
To get \( \frac{dy}{dx} \) by itself:
\( \frac{dy}{dx} = \frac{-(2x + y)}{x + 2y} \)
In simple words: We take the derivative of each part, recalling the product rule for \(xy\). Then we move terms to solve for \( \frac{dy}{dx} \).

Exam Tip: Constants like 100 differentiate to 0. Always be sure to combine like terms correctly before isolating \( \frac{dy}{dx} \).

Question 6. x³ + x²y + xy² + y³ = 81
Answer: We perform implicit differentiation on the given equation with respect to \(x\).
\(x^3 + x^2y + xy^2 + y^3 = 81\)
Differentiating each term with respect to \(x\):
\( \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2y) + \frac{d}{dx}(xy^2) + \frac{d}{dx}(y^3) = \frac{d}{dx}(81) \)
\( 3x^2 + (2xy + x^2\frac{dy}{dx}) + (y^2 + x \cdot 2y \frac{dy}{dx}) + 3y^2\frac{dy}{dx} = 0 \)
Now, we group the terms that contain \( \frac{dy}{dx} \):
\( 3x^2 + 2xy + y^2 + (x^2 + 2xy + 3y^2)\frac{dy}{dx} = 0 \)
Move terms without \( \frac{dy}{dx} \) to the other side:
\( (x^2 + 2xy + 3y^2)\frac{dy}{dx} = -3x^2 - 2xy - y^2 \)
Finally, we solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{-(3x^2 + 2xy + y^2)}{x^2 + 2xy + 3y^2} \)
In simple words: Differentiate each part of the equation, using the product rule for terms like \(x^2y\) and \(xy^2\). Group all \( \frac{dy}{dx} \) terms, then solve for it.

Exam Tip: Pay close attention to the product rule application for \(x^2y\) and \(xy^2\), and ensure proper use of the chain rule for \(y^2\) and \(y^3\).

Question 7. sin² y + cos xy = π
Answer: We will differentiate this implicit equation with respect to \(x\).
\( \sin^2 y + \cos xy = \pi \)
Differentiating each term with respect to \(x\):
\( \frac{d}{dx}(\sin^2 y) + \frac{d}{dx}(\cos xy) = \frac{d}{dx}(\pi) \)
\( 2\sin y \cos y \frac{dy}{dx} - \sin(xy) \left(1 \cdot y + x \frac{dy}{dx}\right) = 0 \)
\( 2\sin y \cos y \frac{dy}{dx} - y\sin(xy) - x\sin(xy)\frac{dy}{dx} = 0 \)
Group the terms that contain \( \frac{dy}{dx} \):
\( (2\sin y \cos y - x\sin(xy))\frac{dy}{dx} = y\sin(xy) \)
Recall that \(2\sin y \cos y = \sin 2y\). Substitute this:
\( (\sin 2y - x\sin(xy))\frac{dy}{dx} = y\sin(xy) \)
Finally, solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{y\sin(xy)}{\sin 2y - x\sin(xy)} \)
In simple words: Differentiate both sides, using chain rule for \( \sin^2 y \) and product rule for \( \cos xy \). Gather terms with \( \frac{dy}{dx} \) and simplify.

Exam Tip: The derivative of \( \sin^2 y \) is \( 2\sin y \cos y \frac{dy}{dx} \). Be careful with the product rule application within the \( \cos xy \) term.

Question 8. sin² x + cos² y = 1
Answer: We apply implicit differentiation with respect to \(x\) to the equation.
\( \sin^2 x + \cos^2 y = 1 \)
Differentiating both sides with respect to \(x\):
\( \frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(\cos^2 y) = \frac{d}{dx}(1) \)
\( 2\sin x \cos x + 2\cos y (-\sin y) \frac{dy}{dx} = 0 \)
\( 2\sin x \cos x - 2\sin y \cos y \frac{dy}{dx} = 0 \)
Recall the double angle identities: \(2\sin x \cos x = \sin 2x\) and \(2\sin y \cos y = \sin 2y\). Substitute these:
\( \sin 2x - \sin 2y \frac{dy}{dx} = 0 \)
Now, we rearrange to solve for \( \frac{dy}{dx} \):
\( \sin 2x = \sin 2y \frac{dy}{dx} \)
\( \frac{dy}{dx} = \frac{\sin 2x}{\sin 2y} \)
In simple words: We differentiate each term using chain rule and trigonometric identities. Then, we rearrange to find \( \frac{dy}{dx} \).

Exam Tip: Recognizing and applying double angle identities (e.g., \( 2\sin A \cos A = \sin 2A \)) can greatly simplify the expressions.

Question 9. y = sin⁻¹\left(\frac{2 x}{1+x^{2}}\right)
Answer: To simplify the differentiation, we use a substitution.
Let \( x = \tan \theta \). Then \( \theta = \tan^{-1} x \).
Substitute \(x = \tan \theta\) into the equation:
\( y = \sin^{-1}\left(\frac{2 \tan \theta}{1+\tan^{2} \theta}\right) \)
We know the trigonometric identity: \( \frac{2 \tan \theta}{1+\tan^{2} \theta} = \sin 2\theta \).
So, the equation becomes:
\( y = \sin^{-1}(\sin 2\theta) \)
\( y = 2\theta \)
Now, substitute back \( \theta = \tan^{-1} x \):
\( y = 2\tan^{-1} x \)
Finally, differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}(2\tan^{-1} x) \)
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} \)
\( \frac{dy}{dx} = \frac{2}{1+x^2} \)
In simple words: We make a substitution with \( \tan \theta \) to simplify the expression. After simplifying using trigonometric identities, we replace \( \theta \) back with \( \tan^{-1} x \) and then differentiate.

Exam Tip: For inverse trigonometric functions, substitution (e.g., \(x = \sin \theta\), \(x = \tan \theta\), \(x = \cos \theta\)) is often the easiest path to differentiation. Remember standard trigonometric identities.

Question 10. y = tan⁻¹\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)
Answer: To simplify this expression before differentiating, we use a trigonometric substitution.
Let \( x = \tan \theta \). Then \( \theta = \tan^{-1} x \).
Substitute \(x = \tan \theta\) into the given equation:
\( y = \tan^{-1}\left(\frac{3 \tan \theta-\tan^{3} \theta}{1-3 \tan^{2} \theta}\right) \)
We recognize the triple angle identity for tangent: \( \frac{3 \tan \theta-\tan^{3} \theta}{1-3 \tan^{2} \theta} = \tan 3\theta \).
So, the equation simplifies to:
\( y = \tan^{-1}(\tan 3\theta) \)
\( y = 3\theta \)
Substitute back \( \theta = \tan^{-1} x \):
\( y = 3\tan^{-1} x \)
Now, differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}(3\tan^{-1} x) \)
\( \frac{dy}{dx} = 3 \cdot \frac{1}{1+x^2} \)
\( \frac{dy}{dx} = \frac{3}{1+x^2} \)
In simple words: We use \( x = \tan \theta \) to simplify the inner part of the inverse tangent. By using a trigonometric identity, the expression becomes much simpler, allowing for easy differentiation after substituting back.

Exam Tip: Be familiar with triple angle formulas like \( \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} \) as they are commonly used for simplifying inverse trigonometric functions.

Question 11. y = cos⁻¹\left(\frac{1-x^{2}}{1+x^{2}}\right)
Answer: We use a substitution to make the differentiation easier.
Let \( x = \tan \theta \). Then \( \theta = \tan^{-1} x \).
Substitute \(x = \tan \theta\) into the equation:
\( y = \cos^{-1}\left(\frac{1-\tan^{2} \theta}{1+\tan^{2} \theta}\right) \)
We know the double angle identity: \( \frac{1-\tan^{2} \theta}{1+\tan^{2} \theta} = \cos 2\theta \).
So, the equation becomes:
\( y = \cos^{-1}(\cos 2\theta) \)
\( y = 2\theta \)
Now, substitute back \( \theta = \tan^{-1} x \):
\( y = 2\tan^{-1} x \)
Finally, differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}(2\tan^{-1} x) \)
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} \)
\( \frac{dy}{dx} = \frac{2}{1+x^2} \)
In simple words: Substituting \( x = \tan \theta \) helps to simplify the function into a basic double-angle cosine identity. After returning to \(x\) terms, we differentiate the simpler expression.

Exam Tip: Recognize the identity \( \cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta} \). This is a common form for simplification in inverse trigonometric problems.

Question 12. y = sin⁻¹\left(\frac{1-x^{2}}{1+x^{2}}\right), 0 < x < 1
Answer: We use a substitution to simplify the expression for differentiation.
Let \( x = \tan \theta \). Then \( \theta = \tan^{-1} x \).
Substitute \(x = \tan \theta\) into the equation:
\( y = \sin^{-1}\left(\frac{1-\tan^{2} \theta}{1+\tan^{2} \theta}\right) \)
We know the identity: \( \frac{1-\tan^{2} \theta}{1+\tan^{2} \theta} = \cos 2\theta \).
So, the equation becomes:
\( y = \sin^{-1}(\cos 2\theta) \)
To solve \( \sin^{-1}(\cos A) \), we use the identity \( \cos A = \sin(\frac{\pi}{2} - A) \):
\( y = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - 2\theta\right)\right) \)
\( y = \frac{\pi}{2} - 2\theta \)
Now, substitute back \( \theta = \tan^{-1} x \):
\( y = \frac{\pi}{2} - 2\tan^{-1} x \)
Finally, differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - 2\tan^{-1} x\right) \)
\( \frac{dy}{dx} = 0 - 2 \cdot \frac{1}{1+x^2} \)
\( \frac{dy}{dx} = \frac{-2}{1+x^2} \)
In simple words: We substitute \(x\) with \( \tan \theta \) and use trigonometric identities to simplify the expression. After converting \( \cos 2\theta \) to a sine form, we get a simpler equation to differentiate.

Exam Tip: When you have \( \sin^{-1}(\cos A) \) or \( \cos^{-1}(\sin A) \), use the complementary angle identities (e.g., \( \cos A = \sin(\frac{\pi}{2} - A) \)) to simplify the expression.

Question 13. y = cos⁻¹\left(\frac{2 x}{1+x^{2}}\right), − 1 < x < 1
Answer: We begin by using a substitution to simplify the function.
Let \( x = \tan \theta \). Then \( \theta = \tan^{-1} x \).
Substitute \(x = \tan \theta\) into the equation:
\( y = \cos^{-1}\left(\frac{2 \tan \theta}{1+\tan^{2} \theta}\right) \)
We know the identity: \( \frac{2 \tan \theta}{1+\tan^{2} \theta} = \sin 2\theta \).
So, the equation becomes:
\( y = \cos^{-1}(\sin 2\theta) \)
To solve \( \cos^{-1}(\sin A) \), we use the identity \( \sin A = \cos(\frac{\pi}{2} - A) \):
\( y = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - 2\theta\right)\right) \)
\( y = \frac{\pi}{2} - 2\theta \)
Now, substitute back \( \theta = \tan^{-1} x \):
\( y = \frac{\pi}{2} - 2\tan^{-1} x \)
Finally, differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - 2\tan^{-1} x\right) \)
\( \frac{dy}{dx} = 0 - 2 \cdot \frac{1}{1+x^2} \)
\( \frac{dy}{dx} = \frac{-2}{1+x^2} \)
In simple words: We simplify the function by substituting \( x = \tan \theta \). Using trigonometric identities, we change the sine term to a cosine term, which then simplifies to a linear function of \( \theta \). After substituting back, we differentiate.

Exam Tip: When dealing with \( \cos^{-1}(\sin A) \), convert \( \sin A \) to \( \cos(\frac{\pi}{2} - A) \) to simplify the inverse cosine function effectively.

Question 14. y = sin⁻¹\left(2 x \sqrt{1-x^{2}}\right), – \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}
Answer: We use a substitution to simplify the expression before differentiation.
Let \( x = \sin \theta \). Then \( \theta = \sin^{-1} x \).
Substitute \(x = \sin \theta\) into the equation:
\( y = \sin^{-1}(2 \sin \theta \sqrt{1-\sin^{2} \theta}) \)
We know that \( \sqrt{1-\sin^{2} \theta} = \sqrt{\cos^{2} \theta} = |\cos \theta| \).
Given the range \( – \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \), which means \( -\frac{\pi}{4} < \theta < \frac{\pi}{4} \), \( \cos \theta \) is positive, so \( |\cos \theta| = \cos \theta \).
Therefore, the expression becomes:
\( y = \sin^{-1}(2 \sin \theta \cos \theta) \)
Using the double angle identity \( 2 \sin \theta \cos \theta = \sin 2\theta \):
\( y = \sin^{-1}(\sin 2\theta) \)
\( y = 2\theta \)
Now, substitute back \( \theta = \sin^{-1} x \):
\( y = 2\sin^{-1} x \)
Finally, differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}(2\sin^{-1} x) \)
\( \frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{1-x^2}} \)
\( \frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}} \)
In simple words: We simplify the function by using the substitution \( x = \sin \theta \). This helps to change the expression into a double angle identity, which then simplifies to a linear function of \( \theta \). After substituting back, we differentiate the simpler term.

Exam Tip: For expressions involving \( \sqrt{1-x^2} \), substitution with \( x = \sin \theta \) or \( x = \cos \theta \) is often very effective. Remember to check the range to correctly handle square roots of squared terms.

Question 15. y = sec⁻¹\left(\frac{1}{2 x^{2}-1}\right), 0 < x < \frac{1}{\sqrt{2}}
Answer: To simplify this expression, we use a trigonometric substitution.
Let \( x = \cos \theta \). Then \( \theta = \cos^{-1} x \).
Substitute \(x = \cos \theta\) into the equation:
\( y = \sec^{-1}\left(\frac{1}{2 \cos^{2} \theta-1}\right) \)
We know the double angle identity: \( 2\cos^{2} \theta - 1 = \cos 2\theta \).
So, the expression becomes:
\( y = \sec^{-1}\left(\frac{1}{\cos 2\theta}\right) \)
Since \( \frac{1}{\cos 2\theta} = \sec 2\theta \), we have:
\( y = \sec^{-1}(\sec 2\theta) \)
\( y = 2\theta \)
Now, substitute back \( \theta = \cos^{-1} x \):
\( y = 2\cos^{-1} x \)
Finally, differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}(2\cos^{-1} x) \)
\( \frac{dy}{dx} = 2 \cdot \left(\frac{-1}{\sqrt{1-x^2}}\right) \)
\( \frac{dy}{dx} = \frac{-2}{\sqrt{1-x^2}} \)
In simple words: We use the substitution \( x = \cos \theta \) to simplify the inner part of the inverse secant. This transforms the expression into a simpler trigonometric form using a double-angle identity, which then allows for direct differentiation.

Exam Tip: For inverse secant or inverse cosine functions with expressions like \( 2x^2-1 \), the substitution \( x = \cos \theta \) is typically very useful because \( 2\cos^2 \theta - 1 = \cos 2\theta \).