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Find an equation of the tangent line to the curve at the point (1, 1).
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\[y=\ln(x e^{x^5})\]
Find the derivative with respect to x first.
So there you go. Now just put in your x value of 1 to figure out the slope.
So that give you your slope, m. Now solve the equation 1 = m*1 + b for b. The equation for the line is y = m*x + b
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y=ln(x)+ln(e^(x^5)) y=ln(x)+x^5 dy/dx=1/x+5x^4
dy/dx = i/x +5x^4 at point(1,1) slope m=1/1 + 5(1)^4=6 y-y1=m(x-xi) y-1=6(x-1) y=6x-6+1 y=6x-5 ans......eq. of the tangent line at pt(1,1)
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