differentiate y with respect to x where y = e^(cot^-1 x^4)^7 2
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we use the chain rule to do the differentiation.
![y = e^{(Cot^{-1}x^4)^7}\\\\Let\ w=Cot^{-1}x,\ \ cot\ w=x\\\frac{1}{x}=tan\ w,\ \ -1/x^2=sec^2w\ w'\\\\w'=-\frac{1}{1+x^2}\\\\d(cot^{-1}x^4)/dx = -\frac{1}{1+x^8}*(4x^3)\\\\\frac{d(e^{z^7})}{dz}=e^{z^7}*7z^6\\\\now\ y'=y*7(cot^{-1})^6*[- \frac{1}{1+x^8}*(4x^3) ]](https://tex.z-dn.net/?f=y+%3D+e%5E%7B%28Cot%5E%7B-1%7Dx%5E4%29%5E7%7D%5C%5C%5C%5CLet%5C+w%3DCot%5E%7B-1%7Dx%2C%5C+%5C+cot%5C+w%3Dx%5C%5C%5Cfrac%7B1%7D%7Bx%7D%3Dtan%5C+w%2C%5C+%5C+-1%2Fx%5E2%3Dsec%5E2w%5C+w%27%5C%5C%5C%5Cw%27%3D-%5Cfrac%7B1%7D%7B1%2Bx%5E2%7D%5C%5C%5C%5Cd%28cot%5E%7B-1%7Dx%5E4%29%2Fdx+%3D+-%5Cfrac%7B1%7D%7B1%2Bx%5E8%7D%2A%284x%5E3%29%5C%5C%5C%5C%5Cfrac%7Bd%28e%5E%7Bz%5E7%7D%29%7D%7Bdz%7D%3De%5E%7Bz%5E7%7D%2A7z%5E6%5C%5C%5C%5Cnow%5C+y%27%3Dy%2A7%28cot%5E%7B-1%7D%29%5E6%2A%5B-+%5Cfrac%7B1%7D%7B1%2Bx%5E8%7D%2A%284x%5E3%29+%5D "y = e^{(Cot^{-1}x^4)^7}\\\\Let\ w=Cot^{-1}x,\ \ cot\ w=x\\\frac{1}{x}=tan\ w,\ \ -1/x^2=sec^2w\ w'\\\\w'=-\frac{1}{1+x^2}\\\\d(cot^{-1}x^4)/dx = -\frac{1}{1+x^8}*(4x^3)\\\\\frac{d(e^{z^7})}{dz}=e^{z^7}*7z^6\\\\now\ y'=y*7(cot^{-1})^6*[- \frac{1}{1+x^8}*(4x^3) ]")
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