Find the gradient of the curve y = sin(2x) + 3 at the point where x = pi

This is a question that relates to the topic of differentiation. Typically students encounter this topic near the end of the first term of the first year of their A Level.The equation of the curve, y = sin(2x) +3 needs to be differentiated to find the general gradient of the line.
This uses the chain rule which is [f(g(x))]' = f'(g(x))g'(x)
The equation is, therefore, differentiated in two parts.[sin(2x)]' = cos(2x) x 2 = 2cos(2x)[+ 3]' = 0
Therefore the general gradient of the curve is: d/dx = 2cos(2x)
The question asks for a specific gradient, at point when x=pi. Therefore, we substitute the value of x=pi into the equation.
d/dx = 2cos(2pi) = 2 x 1 = 2
Therefore, the gradient of the curve at the point where x=pi is 2

AS
Answered by Agnieszka S. Maths tutor

5409 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Differentiate Sin^2(X) with respect to X


For a curve of gradient dy/dx = (2/(x^2))-x/4, determine a) d^2y/dx^2 b) the stationary point where y=5/2 c) whether this is a maximum or minmum point and d) the equation of the curve


Find dy/dx where y= x^3(sin(x))


Solve the equation 5^x = 8, giving your answer to 3 significant figures.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning