find the equation of the tangent line to the curve ylnxex2 a

Solution

We have given the function as y=ln(xe^x^2) we know that tangent line equation = y = mx + b we know that derivative of the function gives the slope of the tangent. So, first we need to find the slope value. y=ln(xe^x^2) y\' = (1/(xe^x^2))*d/dx((xe^x^2)) y\' = (2x^2 + 1)/x now, we will plug the value of x = 1, we have given to get the value of slope. we will get slope m = (2(1)^2 + 1)/1 = (2 + 1)/1 = 3 now, we will use the slope intercept form to get the value of b y = mx + b we have the values of y = 1, m = 3and x = 1 plugging all the value we will get 1 = 3*1 + b b = -2 now, just plug the value of m and b into the slope intercept form to get the tangent line equation we will get y = 3x - 2 Hope this will help you!