Differentiate how? Part 4

Calculus Level 2

The value of d ( x x ) d x \frac { d\left( { x }^{ x } \right) }{ dx } is:

x x log ⁡ x { x }^{ x }\log { x } None of these x x ( log ⁡ e x + 1 ) { x }^{ x }(\log _{ e }{ x } + 1 ) x . x x − 1 x.{ x }^{ x-1 }

Dinesh Nath Goswami by Dinesh Nath Goswami , 20, India

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2 solutions

Mrinmay Dhar
Aug 15, 2015

x x = y x ⋅ l o g ( x ) = l o g ( y ) x ⋅ 1 x + l o g ( x ) = 1 y d y d x d y d x = y ⋅ ( 1 + l o g ( x ) ) ⇒ x x ( 1 + l o g ( x ) ) { x }^{ x }=y\\ x\cdot log(x)=log(y)\\ x\cdot \frac { 1 }{ x } +log(x)=\frac { 1 }{ y } \frac { dy }{ dx } \\ \frac { dy }{ dx } =y\cdot (1+log(x))\\ \Rightarrow { x }^{ x }(1+log(x))

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Raj Rajput
Aug 15, 2015

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice hand writing

siddharth bhatt - 5 years, 10 months ago

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thank you :)

RAJ RAJPUT - 5 years, 10 months ago

Why isn't the answer x . x x − 1 x.x^{x-1}

Ahmad Khamis - 5 years, 10 months ago

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because that rule is applicable for x^a where a is a constant not for x^x, in such cases we have to take log on both sides and rest you know

RAJ RAJPUT - 5 years, 10 months ago

x^x=x.x^x-1

Shashwat Avasthi - 5 years, 9 months ago

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