# Implicit differentiation exercise

Try these problems to freshen up your Calculus skills$!$

• If $\log { ({ x }^{ 2 }+{ y }^{ 2 }) = 2\tan ^{ -1 }{ \left( \frac { y }{ x } \right) } }$, then show that $\frac { dy }{ dx } =\frac { x+y }{ x-y }$.

• If $x\sqrt { 1+y } +y\sqrt { 1+x } =0$, then prove that $\frac { dy }{ dx } =\frac { -1 }{ { \left( x+1 \right) }^{ 2 } }$.

• If $\cos ^{ -1 }{ \left( \frac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) } =\tan ^{ -1 }{ a }$, then prove that $\frac { dy }{ dx } =\frac { y }{ x }$.

• If $\sin { y=x\sin { \left( a+y \right) } }$, then prove that $\frac { dy }{ dx } =\frac { \sin ^{ 2 }{ \left( a+y \right) } }{ \sin { a } }$.

• If ${ x }^{ 2 }+{ y }^{ 2 }=t-\frac { 1 }{ t }$ and ${ x }^{ 4 }+{ y }^{ 4 }={ t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } }$, then prove that $\frac { dy }{ dx } =\frac { 1 }{ { x }^{ 3 }y }$.

Do post solutions. Thanks$!$

Swapnil Das Note by Swapnil Das
5 years, 4 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Third one :

$\cos^{-1}\left( \dfrac{x^{2}-y^{2}}{x^{2}+y^{2}} \right) = \tan^{-1}(a)$
$\left( \dfrac{x^{2}-y^{2}}{x^{2}+y^{2}}\right) = \cos(\tan^{-1}(a))$
Apply componendo-dividendo,
$\dfrac{x^{2}}{y^{2}} = -c$ where c is a constant.
This, is just the equation of straight line through the origin.
$\therefore \dfrac{dy}{dx} =(\text{Slope}) = \dfrac{y}{x}$

- 5 years, 4 months ago

Fifth one :
$x^{2} + y^{2} = t - \dfrac{1}{t}$
Squaring,
$x^{4} + y^{4} + 2x^{2}y^{2} = t^{2} + \dfrac{1}{t^{2}} - 2$
$\therefore x^{2}y^{2} = - 1$
$y^{2} = \dfrac{-1}{x^{2}}$
Differentiate ,

$2y \dfrac{dy}{dx} = \dfrac{2}{x^{3}}$
$\therefore \dfrac{dy}{dx} = \dfrac{1}{x^{3}y}$

- 5 years, 4 months ago

First one,
Differentiating,
$\dfrac{2}{x^{2}+y^{2}}\left(x+y\dfrac{dy}{dx}\right) = 2\dfrac{x^{2}}{x^{2}+y^{2}}\left(x\dfrac{dy}{dx} - y\right) \cdot \dfrac{1}{x^{2}}$
$x + y \dfrac{dy}{dx} = x\dfrac{dy}{dx} - y$
$x + y = (x-y)\dfrac{dy}{dx}$
$\dfrac{dy}{dx} = \dfrac{x+y}{x-y}$

- 5 years, 4 months ago

Second one :

$x\sqrt{1+y} + y\sqrt{1+x} = 0$
$x^{2}(1+y) = y^{2}(1+x)$
$x^{2} - y^{2} = y^{2}x - x^{2}y$
$(x+y)(x-y) = xy(y-x)$
$x+y = -xy$
$x = -y(1+x)$
$-y = \dfrac{x}{1+x}$
$-y = 1 - \dfrac{1}{1+x}$
Differentiate,
$-\dfrac{dy}{dx} = -\dfrac{-1}{(1+x)^{2}}$
$\dfrac{dy}{dx} = \dfrac{-1}{(1+x)^{2}}$

- 5 years, 4 months ago

Fourth one :
$\sin(y) = x\sin(a+y)$
Differentiating,
$\cos(y)\dfrac{dy}{dx} = \sin(a+y) + x\cos(a+y)\dfrac{dy}{dx}$
$\dfrac{dy}{dx} = \dfrac{\sin(a+y)}{\cos(y) - x\cos(a+y)}$
$\dfrac{dy}{dx} = \dfrac{\sin(a+y)}{\cos(y) - x(\cos(a)\cos(y) - \sin(a)\sin(y))} = \dfrac{\sin(a+y)}{\cos(y)(1-x\cos(a)) + x\sin(a)\sin(y)}$

Frimt the original equation,
$\sin(y) = x\sin(y)\cos(a) + x\cos(y)\sin(a)$
$1 - x\cos(a) = x\sin(a)\cot(y)$

Substituting,
$\dfrac{dy}{dx} = \dfrac{\sin(a+y)}{\cos(y)x\sin(a)\cot(y) + x\sin(a)\sin(y)}$
$\therefore \dfrac{dy}{dx} = \dfrac{\sin(a+y)\sin(y)}{x\sin(a)\left(\cos^{2}(y)+\sin^{2}(y)\right)}$
Substituting,
$\sin(y) = x\sin(a+y)$
$\dfrac{dy}{dx} = \dfrac{x\sin(a+y)\cdot \sin(a+y)}{x\sin(a)} = \dfrac{\sin^{2}(a+y)}{\sin(a)}$

- 5 years, 4 months ago

@Vighnesh Shenoy Thanks, I am indebted to you.

- 5 years, 4 months ago