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2015-10-31T13:56:33+05:30

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We need to know the differential of cot^-1 x.  first I found that because I dont remember its formula.  differential of  x^4 = 4x^3    and  Ln  y =  1/y

y=e^{(cot^{-1}x^4)^7}\\\\let\ z=cot^{-1}x,\ x=cotz,\ \ dx=-cosec^2z\ dz,\ z'=-\frac{1}{1+x^2}\\\\y'= e^{(cot^{-1}x^4)^7} * 7*(cot^{-1}x^4)^6*(\frac{-1}{1+(x^4)^2}*4x^3)\\\\.

We use the chain rule of differentiation for this.
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y=e^{(cot^{-1}x^4)^7}\\\\ Ln\ y=(cot^{-1}x^4)^7\\\\\frac{1}{y}y'=7(cot^{-1}x^4)6*(\frac{-1}{1+(x^4)^2}*4x^3)\\\\.Now\ find\ y'

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no no.. i will explain in simple terms...
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ok see.
  • Brainly User
2015-10-31T14:05:31+05:30
Y=e^(cot inverse x^4)^7
dy/dx= (e^cot inverse x^4)^7   d/dx (cot inverse x^4)^7 
dy/dx= e^(cot inverse x^4)^7  7(cot inverse x^4)^6 . d/dx  (cot^-1x)
dy/dx= e^(cot inverse x^4)^7 ×7 (cot^-1x^4)^6.  -(1/1+(x^4)^2)  d/dx x^4
dy/dx=e^(cot inverse x^4)^7×7(cot inverse x^4)^6 . (-1/1+(x^4)^2) . 4(x^3) 
this is the differentiation of this function 

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(-1/1+x²) is not right... replace x here by x^4...
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