If y = 2 2 ^x , then show that dy/dx = y( log 2 ) 2 *2 x

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kkbisht · Answer

Let y=2 2^x Take log we obtain logy = 2 x log2 on differetiation wrt x we get (1/y) dy/dx= log 2 .( 2 x Log2)= ( log2) 2 2 x = > dy/dx= y (log) 2 .2 x kkbisht

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If y = 2 2 ^x , then show that dy/dx = y( log 2 ) 2 *2 x

If y = 2²^{^x}, then show that dy/dx = y( log 2 )²*2^x

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If y = 2 2 ^x , then show that dy/dx = y( log 2 ) 2 *2 x

If y = 22^x, then show that dy/dx = y( log 2 )2 *2x

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If y = 2²^{^x}, then show that dy/dx = y( log 2 )²*2^x