What is the derivative of y=sin^3(x)+sin(x^3) ?
What is
3sin^2(x)*cosx + cos(x^3)*3x^2
if one has cos^n (u^P) the derivative is
d/dx (sin^n (z)) where z=u^P then
= n cos^(n-1)(z)*-sin(z)*dz/dx
and dz/dx will be p*u^(p-1) du/dx and du/dx is depends on what u s.
the point is chain the derivative.
A container in the form of a right circular cone (vertex down) has a radius of 4m and height of 16m. If water is poured into the container at the constant rate of 16m^3/min, how fast is the water level rising when the water is 8m deep?
V=(pi/)(r^2)h
ddx(sin(x3)+(sin(x))3)
=ddx(sin(x3))+ddx((sin(x))3)
=3⋅(sin(x))2⋅ddx(sin(x))+cos(x3)⋅ddx(x3)
=3x2⋅cos(x3)+3⋅cos(x)⋅(sin(x))2
To find the derivative of the given function y = sin^3(x) + sin(x^3), we can use the sum rule and chain rule of differentiation.
The sum rule states that the derivative of a sum of functions is the sum of their derivatives. So, we need to find the derivative of each term separately.
Let's start with the first term: y1 = sin^3(x). To find its derivative, we can use the chain rule. The chain rule states that if we have a composition of functions, f(g(x)), then the derivative is given by f'(g(x)) * g'(x).
In this case, the outer function is f(u) = u^3, and the inner function is g(x) = sin(x). The derivative of the outer function f(u) = u^3 with respect to u is f'(u) = 3u^2. The derivative of the inner function g(x) = sin(x) with respect to x is g'(x) = cos(x).
Applying the chain rule, we have:
dy1/dx = f'(g(x)) * g'(x) = 3(sin(x))^2 * cos(x) = 3sin^2(x) * cos(x).
Now let's move on to the second term: y2 = sin(x^3). Using the chain rule again, we have:
dy2/dx = cos(x^3) * (3x^2).
Finally, using the sum rule, we can add the derivatives of y1 and y2 together to find the derivative of the whole function y = sin^3(x) + sin(x^3):
dy/dx = dy1/dx + dy2/dx = 3sin^2(x) * cos(x) + cos(x^3) * (3x^2).
So, the derivative of y = sin^3(x) + sin(x^3) is 3sin^2(x) * cos(x) + cos(x^3) * (3x^2).