Can you differentiate it?

Calculus Level 1

d d x sin ⁡ 2 x + cos ⁡ 2 x x = ? \large\dfrac{d}{dx} \dfrac{\sin^{2} x+\cos^{2} x}{x}=?

Bonus: after answering this,can you also differentiate

sin ⁡ 2 x + cos ⁡ 2 x 2 x ? \dfrac{\sin^{2} x+\cos^{2} x}{2x}?

− 1 x 2 -\dfrac{1}{x^2} 2 x 2 2x^2 1 1 1 x 2 \dfrac{1}{x^2}

Nazmus Sakib by Nazmus sakib , 24, Bangladesh

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2 solutions

Saksham Jain
Nov 18, 2017

As we know s i n 2 x + cos ⁡ 2 x = 1 \large sin^2x+\cos^2x=1 therefore We have to differentiate x − 1 x^{-1} which we can easily do by using n × x n − 1 \large n\times x^{n-1}\ Now bonus question is very easy

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Ananth Jayadev
Nov 21, 2017

Solution to the bonus problem:

We know the following trigonometric identity,

s i n 2 x + cos ⁡ 2 x = 1 \large sin^2x+\cos^2x=1

Thus we really have to do the following task:

d d x 1 2 x \large \frac { d }{ dx } \frac { 1 }{ 2x }

Using either the Power Rule or the Quotient Rule, the derivative becomes − 1 4 x 2 \large -\frac { 1 }{ { 4x }^{ 2 } }

Edit: I meant − 1 2 x 2 \large -\frac { 1 }{ { 2x }^{ 2 } }

0 Helpful 0 Interesting 0 Brilliant 0 Confused

How you can say d d x 1 2 x = − 1 4 x 2 \dfrac{d}{dx} \dfrac{1}{2x}=-\dfrac{1}{4x^{2}} ?

Nazmus sakib - 3 years, 6 months ago

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See my edit.

Ananth Jayadev - 3 years, 6 months ago

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