Using differentiation find the equation of the tangent to the curve y= 4 + ln(x+1) at the point where x=0 Thanks!
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Originally Posted by Oasis1993 Using differentiation find the equation of the tangent to the curve y= 4 + ln(x+1) at the point where x=0 Thanks! Have you found the derivative of 4+ ln(x+1)?
i think im doing the mistake there. is the derivative of y= 4+ ln(x+1) 1/(x+1) or 4/ (x+1) ?
Originally Posted by Oasis1993 i think im doing the mistake there. is the derivative of y= 4+ ln(x+1) 1/(x+1) or 4/ (x+1) ? Neither. The derivative of "4" is "0" and the derivative of "ln(x+1)" is " ". The derivative of the sum f+ g is f'+ g' so the derivative of 4+ ln(x+1) is .
yes the derivative of the equation is 1 / (x+1). But what do i do to get the equation of the tangent? I put 0 instead of x which gives 1?..and then?
The first thing you should have learned about the derivative is that it generalizes the slope of a line. The slope of the tangent line to a curve at a point is the derivative at that point.
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