Question

use implicit differentiation to find the slope of the tangent line at (2,1) and use it to find the equation of the tangent line in y=mx + b form

Answer 1

heres the equation \[3^x+\log_2_(xy)=10\]

Answer 2

ok sorry one more time \[3^x+\log_2(xy)=10\] thats the right equation

Answer 3

and that is log base 2 if you can't read that

Answer 4

\[( Ln2*2^{x} + 1/xyLna *y) + ( 1/xyLna *x)y' = 0\]
that would be the implicit derivative

Answer 5

i have no clue how you did that haha

Answer 6

now: express y' from this expression and you got the slope

Answer 7

take partial derivatives, :)

Answer 8

i hope i didnt make any mistakes in derivation, make sure by ur self

Answer 9

\[d/dx[a ^{x}] =Ln(a)*a ^{x}\]

Answer 10

the rest of derivatives are simple

Answer 11

\[ d/dx[\log_{a} x]=1/xLn(a)\]

Answer 12

got it?

Answer 13

Can you do the differentiation?

I would change \( 3^x \) to \( e^{x ln3} \)
its derivative is \( ln 3 e^{x ln3} = ln(3) 3^{x}\)

Answer 14

rewrite log_2(xy) as
\[ \frac{1}{ln(2)} ln(xy) \]
its derivative is
\[ \frac{1}{ln(2)} \frac{1}{xy} (x dy/dx + y dx/dx) \]
or just
\[ \frac{1}{ln(2)} \frac{1}{xy} (x \frac{dy}{dx} + y ) \]

Answer 15

multiply through by ln(2)xy:
\[ y ln(2)ln(3) x 3^x + y + x \frac{dy}{dx} = 0 \]
and
\[ \frac{dy}{dx} =-\frac{y}{x}(ln(2)ln(3)x3^x+1) \]