Find the derivative of y =Cos(x2) with respect to x and the derivative of y=sin^2 (4x)cos(3x)with respect to x

Answer 1

y =Cos(x^2)

just write down answer;

y'= -(2x)sin(x^2)

y=sin^2 (4x)cos(3x)


d(sin4x)/dx=cos4x
d(sin4x*sin4x)/dx
=4sin4x*cos4x
+4cos4x*sin4x
=8sin4x*cos4x

d(cos3x)/dx= -3sin3x

using d(uv)=udv+vdu

d{(sin4x)^2*cos3x}/dx
= -3sin3x*(sin4x)^2
+8sin4x*cos4x*cos3x
=sin4x(8cos4x*cos3x
-3sin3x*sin4x)

[note;just write down the
answer to the first part of
this question]

i hope that this helps

Answer 2

Use product rule. y=2(4)sin(4x)cos(4x)cos(3x)
-3sin^2(4x)sin(3x)

can still simplify the terms. Hope u can continue it yrself. All the best to yr study...

Answer 3

y=sin^2(4x)cos(3x)

Use the product rule, mixed with the chain rule.

Answer 4

Question 1
y = cos (x²)
Let u = x²
du/dx = 2x
y = cos u
dy/du = - sin u
dy / dx = (dy/du).(du/dx)
dy / dx = - sin (x²) . 2x
dy / dx = - 2x.sin(x²)

Question 2
y = sin²(4x).cos(3x)
dy/dx = 8 sin(4x).cos(4x).cos(3x) - 3.sin(3x).sin²(4x)
= sin(4x).[ 8 cos(4x).cos(3x) - 3.sin(3x).sin(4x) ]

Answer 5

dy/dx for cos(x square) = -sinx * 2x