Jed Patrick P.

asked • 11/04/19# Please help me to solve this equation and it's all about Derivatives (Basic Calculus)

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

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

## 2 Answers By Expert Tutors

Let u = x^{3}+log_{2}x then the problem becomes y=sin^{3}(u) .... dy/dx=3sin^{2}ucos(u)du/dx (chain rule) ... Now du/dx = 3x^{2} + d/dx(log_{2}x). Use the change of base property of logarithms (log2x = ln x/ln2) to find du/dx = 3x^{2}+1/xln2. Reverse the substitution and get dy/dx = 3sin^{2}(x^{3}+log_{2}x)cos(x^{3}+log_{2}x)(3x^{2}+1/xln2)

Jed Patrick P.

THANK YOU, SIR!!!11/07/19

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 (3x^{3})

y' = cos(3x^{3}) * 9x^{2}

In your problem, u = x^{3 }+ log_{2 }x

find u'

then dy/dx = 3*sin^{2 }(x^{3 }+ log_{2 }x) * u'

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Sidney D.

11/04/19