Please help me to solve this equation and it's all about Derivatives (Basic Calculus)

Question

This is a performance task and I don't know what to do here 😭

y=sin^3(x^3+log2 X)

Answer 1

Let u = x³+log₂x then the problem becomes y=sin³(u) .... dy/dx=3sin²ucos(u)du/dx (chain rule) ... Now du/dx = 3x² + d/dx(log₂x). Use the change of base property of logarithms (log2x = ln x/ln2) to find du/dx = 3x²+1/xln2. Reverse the substitution and get dy/dx = 3sin²(x³+log₂x)cos(x³+log₂x)(3x²+1/xln2)

Answer 2

I assume you are being asked to take the derivative dy/dx. If not, then what is the question?

For this, you will use the chain rule: If u is a function of x, then the derivative of f(u) is:

d/dx(f(u)) = df/du * du/dx

Simple example: y = sin (3x³)

y' = cos(3x³) * 9x²

In your problem, u = x³ + log₂ x

find u'

then dy/dx = 3*sin² (x³ + log₂ x) * u'