 Hello friends and how are you all doing today? The question says, find the equation of the tangent and normal to the given curves at the indicated points. Here we have y is equal to x cube at 1 1. So we are given in the question the equation of the curve as y is equal to x cube. First of all we will be finding out the derivative of this function. Differentiating y with respect to x we get dy by dx is equal to 3x square. Now at point 1 comma 1 we have its value as 3 1 square which is equal to 3. This means we have slope of tangent as 3 and slope of normal will be y minus 1 upon dy by dx that is 3. Now with the help of these slopes we can find out equation of both of them. Equation of tangent as well as equation of normal. This is y minus y1 equal to slope into x minus x1. Here also y minus y1 equal to slope of the normal that is minus 1 upon 3 x minus x1. y minus 1 equal to 3x minus 3 which can be further written as y is equal to 3x minus 3 plus 1 is minus 2. Here we have 3y minus 3 plus x minus 1 equal to 0 which further gives us x plus 3y minus 4 equal to 0. So the answer to this part is that is the third part. We have equation of the tangent as y equal to 3x minus 2 and normal as x plus 3y minus 4 equal to 0. This completes the session. Hope you understood it well and enjoyed it too. Have a nice day.