Given that y = ((4x+3)^5)(sin2x), find dy/dx
First of all, we have to use the product rule, since two things are multiplied together. The product rule states that d/dx (u*v) = vu' + uv'
Let u = (4x+3)5 v = sin(2x)
Now, to find u' and v' we have to find du/dx and dv/dx. As we see, we need to use the chain rule to find du/dx
u' = 20(4x+3)4 v' = 2cos(2x)
Finally, dy/dx = vu' + uv' = 20sin(2x)(4x+3)4 + 2cos(2x)(4x+3)5
JN
Related Maths A Level answers
All answers ▸How do I find the area bounded by the curve y=-x^2+4 and the line y=-x+2?
A level Maths question - The graph of y=2sin(2x)+1 is rotated 360 degrees about the x-axis to form a solid. Find the volume enclosed by the curve, the co-ordinate axes and the line x=pi/2
Find the nature of the turning points of the graph given by the equation x^4 +(8/3)*x^3 -2x^2 -8x +177 (6 marks)
Differentiate the equation 4x^5 + 2x^3 - x + 2