Given that y = (sin(6x))(sec(2x) ), find dy/dx
We can find dy/dx by using the product rule: If y=uv then dy/dx = u (dv/dx)+ v (du/dx). In this question u= sin(6x) and v= sec(2x).So du/dx= 6cos(6x) and dv/dx=2sec(2x)tan(2x), using our rules for differentiating trig functions.Subbing this into our product rule formula gives us: dy/dx= sin(6x)(2sec(2x)tan(2x)) + sec(2x)(6cos(6x)).So dy/dx = 2sin(6x)sec(2x)tan(2x) + 6cos(6x)sec(2x), and this is our final answer as it cannot be simplified any more.
EH
Related Maths A Level answers
All answers ▸Find the curve whose gradient is given by dy/dx=xy and which passes through the point (0,3)
How do I integrate 4x*exp(x^2 - 1) with respect to x?
Find dy/dx of 5x^2 + 2y^3 +8 =17.
What is the equation of the curve that has gradient dy/dx=(4x-5) and passes through the point (3,7)?