find dy/dx. y=x2log2(3−2x)

Question

jhonyy9 · Answer

x2 so here this 2 is exponent ?
and the second 2 wan being the base of logarithm ?

kasep10 · Answer

Have you tried chain rule?$$y=x^2 \log_2 (3-2x)$$

itsme.caro · Answer

i understand the chain rule, i get stuck in the log part, not sure how to derive this.

3mar · Answer

May I help?

3mar · Answer

Differentiation of logarithmic function.
log and ln
Feel free to discuss wherever you need

itsme.caro · Answer

ok let me see if i can figure it out

3mar · Answer

ok

itsme.caro · Answer

no i can't

itsme.caro · Answer

i have no idea where to go with this

3mar · Answer

Is this the whole function?
$$y=x^2\log_2(3-2x)$$

3mar · Answer

@itsme.caro

3mar · Answer

Do you follow?

itsme.caro · Answer

what do you mean?

3mar · Answer

Is what I wrote correct?

itsme.caro · Answer

yes, that part is correct.  i thought about using productivity rule, but i don't know how to differentiate log

itsme.caro · Answer

ok so so far i got this: 2xlogbase2 (3-2x)  that part is right... but the second part of the productivity rule is f g'  which would be x^2 (this part i don't know)

3mar · Answer

productivity rule
$$\frac{ d }{ dx }[f(x).g(x)]=f(x).g'(x)+f'(x).g(x)$$


So it will be like that:
$$=(x^2).[(\log_2e).\frac{ 1 }{ 3-2x }]+\log_2(3-2x).[2x]$$

I think it is clear now.

3mar · Answer

Did you get it? @itsme.caro

itsme.caro · Answer

so when you get the derivative the e shows up?

karim728 · Answer

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