A curve C has the equation y=5sin3x + 2cos3x, find the equation of the tangent to the curve at the point (0,2)

Firstly, you need to differentiate the curve y. Using the equation dy/dx of sinkx= kcoskx, you need to differentiate 5sin3x first. So, as k=3 in this example, the answer is 5(3cos3x) which is equal to 15cos3x. If we then look at the 2cos3x, we now have to use the equation: dy/dx of coskx= -ksinkx; k= 3 in this example too so the answer would be 2(-3sin3x) which is equal to -6sin3x. So, if we put that together, dy/dx= 15cos3x - 6sin3x. Now, we have to find the equation of the tangent of the curve at the point (0,2). The general equation of a line is y= mx + c. We will start by finding m which is the gradient. To do this we must substitute the x value (0) into the differentiation that we just calculated. So this would be 15cos3(0) - 6sin3(0). This is equal to 15 which is our gradient (m). To find c we must now substitute the x, y and m values into the y= mx + c equation: 2= 15(0) + c, therefore c is equal to 2. If we now substitute our m and c values into the general equation y= mx + c, the answer to the question is y= 15x + 2.