anonymous
  • anonymous
Find an equation of the tangent line to the curve at the point (1, 1). ln(xe^(x^2)) first I applied the chain rule to get 1/(xe^(x^2)) *d/dx(xe^(x^2)). this is where I get stuck. which exponent do I start with the powers rule on to find this derivative?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I'll tell you you're making it harder than you need to. I'll do it using the chain rule and then I'll show the easy way. You need to do the product rule on the part you were stuck on. If y = ln(xe^(x^2)), then \[dy/dx = 1/xe ^{x^2}[1e ^{x^2}+x*e ^{x^2}2x]\] \[dy/dx = e ^{x^2}(1 + 2x^2)/xe ^{x^2}=(1 + 2x^2)/x\] Easy way - use logarithmic transformation: \[y = \ln(xe ^{x^2})=lnx + lne ^{x^2}= \ln x + x^2\] \[dy/dx = 1/x + 2x\]
anonymous
  • anonymous
thanks I appreciate your help

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