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# find dy/dxy= (e)^x + (10)^x + (x)^x

Rinkoo Gupta
6 years ago

Let y1=xx

Taking log on both sides, get

Logy1=xlogx

Diff w. r. to x ,get

1/y1. Dy1/dx=x.1/x+logx.1

Dy1/dx=y1[1+logx]

=xx[1+logx]

10^x = e^(x ln 10)

Applying the chain rule:dy/dx = e^(x ln 10) * ln 10 = 10^x ln 10

thus dy/dx=ex+10xln10 +xx(1+logx)

Thanks & Regards

Rinkoo Gupta

Arun Kumar IIT Delhi
6 years ago
Hello
$\\\frac{d}{dx}\left(e^x+10^x+x^x\right)=e^x+10^x\ln \left(10\right)+x^x\left(\ln \left(x\right)+1\right)\\$
Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Arun Kumar IIT Delhi
6 years ago
Hello
$\\\frac{d}{dx}\left(e^x+10^x+x^x\right)=e^x+10^x\ln \left(10\right)+x^x\left(\ln \left(x\right)+1\right)$
Thanks & Regards
Arun Kumar
Btech, IIT Delhi