Question

Derive

2^xy + x^2 - 3xy = 5

I’m assuming it is implicit differentiation but how

Answer 1

`Well\;\mathrm{treat\:}y\mathrm{\:as\:}y\left(x\right)`

`\mathrm{Differentiate\:both\:sides\:of\:the\:equation\:with\:respect\:to\:}x`

`(d)/(dx)\left(2^xy+x^2-3xy\right)=(d)/(dx)\left(5\right)`


`(d)/(dx)\left(2^xy+x^2-3xy\right) \ \textgreater \ \mathrm{Apply\:the\:Sum/Difference\:Rule}:\ \left(f\pm g\right)^'=f^'\pm g^'`


`(d)/(dx)\left(2^xy\right) \ \textgreater \ \mathrm{Apply\:the\:Product\:Rule}:\ \left(f\cdot g\right)'=f^'\cdot g+f\cdot g^'\`


`f=2^x,\:g=y \ \textgreater \ (d)/(dx)\left(2^x\right)y+(d)/(dx)\left(y\right)2^x`


`(d)/(dx)\left(2^x\right) \ \textgreater \ \mathrm{Apply\:exponent\:rule}:\ \:a^b=e^(b\ln \left(a\right)) \ \textgreater \ 2^x=e^(x\ln \left(2\right))`

`(d)/(dx)\left(e^(x\ln \left(2\right))\right) \ \textgreater \ \mathrm{Apply\:the\:chain\:rule}:\ (df\left(u\right))/(dx)=(df)/(du)\cdot (du)/(dx)`

`f=e^u,\:\:u=x\ln \left(2\right) \ \textgreater \ (d)/(du)\left(e^u\right)(d)/(dx)\left(x\ln \left(2\right)\right)`


`(d)/(du)\left(e^u\right) \ \textgreater \ \mathrm{Apply\:the\:common\:derivative}:\ (d)/(du)\left(e^u\right)=e^u \ \textgreater \ e^u`


`(d)/(dx)\left(x\ln \left(2\right)\right) \ \textgreater \ \mathrm{Take\:the\:constant\:out}:\ \left(a\cdot f\right)^'=a\cdot f^'`

`\ln \left(2\right)(d)/(dx)\left(x\right) \ \textgreater \ \mathrm{Apply\:the\:common\:derivative}:\ (d)/(dx)\left(x\right)=1 \ \textgreater \ \ln \left(2\right)\cdot \:1`

`Simplify \ \textgreater \ \ln \left(2\right)`


`e^u\ln \left(2\right) \ \textgreater \ \mathrm{Substitute\:back}\:u=x\ln \left(2\right) \ \textgreater \ e^(x\ln \left(2\right))\ln \left(2\right)`


`e^(x\ln \left(2\right)) \ \textgreater \ \mathrm{Apply\:exponent\:rule}:\ \:a^(bc)=\left(a^b\right)^c \ \textgreater \ \left(e^(\ln \left(2\right))\right)^x`

`\mathrm{Apply\:log\:rule}:\ \:a^(\log _a\left(b\right))=b \ \textgreater \ e^(\ln \left(2\right))=2 \ \textgreater \ \ln \left(2\right)\cdot \:2^x`

`\ln \left(2\right)\cdot \:2^xy+(d)/(dx)\left(y\right)2^x`


`(d)/(dx)\left(x^2\right) \ \textgreater \ \mathrm{Apply\:the\:Power\:Rule}:\ (d)/(dx)\left(x^a\right)=a\cdot x^(a-1) \ \textgreater \ 2x^(2-1) \ \textgreater \ 2x`


`(d)/(dx)\left(3xy\right) \ \textgreater \ \mathrm{Take\:the\:constant\:out}:\ \left(a\cdot f\right)^'=a\cdot f^'`

`3(d)/(dx)\left(xy\right) \ \textgreater \ \mathrm{Apply\:the\:Product\:Rule}:\ \left(f\cdot g\right)'=f^'\cdot g+f\cdot g^'`

`f=x,\:g=y \ \textgreater \ 3\left((d)/(dx)\left(x\right)y+(d)/(dx)\left(y\right)x\right)`


`(d)/(dx)\left(x\right) \ \textgreater \ \mathrm{Apply\:the\:common\:derivative}:\ (d)/(dx)\left(x\right)=1`

`3\left(1\cdot \:y+(d)/(dx)\left(y\right)x\right) \ \textgreater \ Simplify \ \textgreater \ 3\left(y+x(d)/(dx)\left(y\right)\right)`


`(d)/(dx)\left(5\right) \ \textgreater \ \mathrm{Derivative\:of\:a\:constant}:\ (d)/(dx)\left(a\right)=0`


`Now\;we\;have \ \textgreater \ \ln \left(2\right)\cdot \:2^xy+(d)/(dx)\left(y\right)2^x+2x-3\left(y+x(d)/(dx)\left(y\right)\right)`


`\mathrm{For\:convenience,\:write\:}(d)/(dx)\left(y\right)\mathrm{\:as\:}y^('\:)`

`\ln \left(2\right)\cdot \:2^xy+y^('\:)\cdot \:2^x+2x-3\left(y+xy^('\:)\right)=0`

Now we have to isolate y^'.


`\ln \left(2\right)\cdot \:2^xy+y^('\:)\cdot \:2^x+2x-3\left(y+xy^('\:)\right)=0`

`\mathrm{Subtract\:}\ln \left(2\right)\cdot \:2^xy+2x\mathrm{\:from\:both\:sides}`
Too big for the formula so i assume you know how to subtract them.
Simplify.


`y^('\:)\cdot \:2^x-3\left(y+xy^('\:)\right)=-\left(\ln \left(2\right)\cdot \:2^xy+2x\right) \ \textgreater \ expand -3\left(y+xy^('\:)\right)`


`\mathrm{Distribute\:parentheses\:using}:\ \:a\left(b+c\right)=ab+ac`

`where\; a=-3,\:b=y,\:c=xy^('\:)`


`-3\cdot \:y-3\cdot \:xy^('\:) \ \textgreater \ y^('\:)\cdot \:2^x-3y-3xy^('\:)=-\left(\ln \left(2\right)\cdot \:2^xy+2x\right)`


`Expand\; -\left(\ln \left(2\right)\cdot \:2^xy+2x\right) \ \textgreater \ \mathrm{Distribute\:parentheses}`

`-\ln \left(2\right)\cdot \:2^xy-2x \ \textgreater \ \mathrm{Apply\:minus-plus\:rules} \ \textgreater \ +\left(-a\right)=-a`

`-\ln \left(2\right)\cdot \:2^xy-2x \ \textgreater \ 2^xy^('\:)-3y-3xy^('\:)=-\ln \left(2\right)\cdot \:2^xy-2x`


`\mathrm{Add\:}3y\mathrm{\:to\:both\:sides}`

`2^xy^('\:)-3y-3xy^('\:)+3y=-\ln \left(2\right)\cdot \:2^xy-2x+3y`

`\mathrm{Simplify} \ \textgreater \ 2^xy^('\:)-3xy^('\:)=-\ln \left(2\right)\cdot \:2^xy-2x+3y`


`2^xy^('\:)-3xy^('\:) \ \textgreater \ \mathrm{Factor\:out\:common\:term\:}y^('\:) \ \textgreater \ y^('\:)\left(-3x+2^x\right)`


`y^('\:)\left(-3x+2^x\right)=-\ln \left(2\right)\cdot \:2^xy-2x+3y`