# How do you differentiate #log_2(x^2sinx^2) #?

first we need to shift the base into natural logs as these are just made for calculus

then we use another standard result namely that

this can be simplified in any number of ways, so i'm going for brevity

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To differentiate ( \log_2(x^2\sin(x^2)) ), use the chain rule and the derivative of the natural logarithm function:

[ \frac{d}{dx}(\log_2(u)) = \frac{1}{u \ln(2)} \cdot \frac{du}{dx} ]

Let ( u = x^2\sin(x^2) ), then differentiate ( u ) with respect to ( x ):

[ \frac{du}{dx} = \frac{d}{dx}(x^2\sin(x^2)) ]

Using the product rule and the chain rule:

[ \frac{du}{dx} = 2x\sin(x^2) + x^2\cos(x^2) \cdot 2x ]

[ \frac{du}{dx} = 2x\sin(x^2) + 2x^3\cos(x^2) ]

Now, substitute ( u ) and ( \frac{du}{dx} ) back into the formula for differentiating ( \log_2(u) ):

[ \frac{d}{dx}(\log_2(x^2\sin(x^2))) = \frac{1}{x^2\sin(x^2)\ln(2)} \cdot \left(2x\sin(x^2) + 2x^3\cos(x^2)\right) ]

[ \frac{d}{dx}(\log_2(x^2\sin(x^2))) = \frac{2x\sin(x^2) + 2x^3\cos(x^2)}{x^2\sin(x^2)\ln(2)} ]

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To differentiate ( \log_2(x^2 \sin(x^2)) ), we can use the chain rule and the properties of logarithms. The derivative of ( \log_b(u) ) with respect to ( x ) is ( \frac{1}{u} \cdot \frac{du}{dx} ).

Let ( u = x^2 \sin(x^2) ). Then, applying the chain rule:

[ \frac{d}{dx}(\log_2(u)) = \frac{1}{u} \cdot \frac{du}{dx} ]

Now, differentiate ( u = x^2 \sin(x^2) ) with respect to ( x ):

[ \frac{du}{dx} = \frac{d}{dx}(x^2 \sin(x^2)) ]

Using the product rule:

[ \frac{du}{dx} = 2x \sin(x^2) + x^2 \cdot \frac{d}{dx}(\sin(x^2)) ]

[ \frac{du}{dx} = 2x \sin(x^2) + x^2 \cdot \cos(x^2) \cdot 2x ]

[ \frac{du}{dx} = 2x \sin(x^2) + 2x^3 \cos(x^2) ]

Now, substitute ( u ) and ( \frac{du}{dx} ) into the expression for the derivative:

[ \frac{d}{dx}(\log_2(x^2 \sin(x^2))) = \frac{1}{x^2 \sin(x^2)} \cdot (2x \sin(x^2) + 2x^3 \cos(x^2)) ]

[ \frac{d}{dx}(\log_2(x^2 \sin(x^2))) = \frac{2x \sin(x^2) + 2x^3 \cos(x^2)}{x^2 \sin(x^2)} ]

So, the derivative of ( \log_2(x^2 \sin(x^2)) ) with respect to ( x ) is ( \frac{2x \sin(x^2) + 2x^3 \cos(x^2)}{x^2 \sin(x^2)} ).

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