Implicit Differentiation

[spynomad]

Hello -

Here is the question that I am stuck on...if sin(x) = ln y and x is greater than zero and less that pi/2, then dy/dx equals...

I started by differentiating both sides and then I got stuck...so basically I got to cos (x) = 1/y dy/dx but I do not know if I even started correctly. Thanks so much : )

 

[BigGlenntheHeavy]

You are already there (almost).

dy/dx = y' = ycos(x) Fin

Check: dy/dx = ycos(x), dy = ycos(x)dx, taking the integral of both sides, we have,

\(\displaystyle \int dy \ = \ \int ycos(x)dx\)

\(\displaystyle \int \frac{dy}{y} \ = \ \int cos(x)dx\)

\(\displaystyle Hence, \ ln(y) \ = \ sin(x)+C, \ C \ = \ 0.\)

 

[spynomad]

That is what I thought but the choices I have do not include that answer. The options include e raised to the power of sin(x) or raised to the power of cos(x). Any ideas??

 

[galactus]

Just take e to both sides of what Glenn gave you. \(\displaystyle y=e^{sin(x)+C}=e^{sin(x)}\cdot e^{C}=C_{1}e^{sin(x)}\)

If C=0, then \(\displaystyle C_{1}=1\) and you have \(\displaystyle e^{sin(x)}\)

 

[Deleted member 4993]

Hello -

Here is the question that I am stuck on...if sin(x) = ln y and x is greater than zero and less that pi/2, then dy/dx equals...

I started by differentiating both sides and then I got stuck...so basically I got to cos (x) = 1/y dy/dx but I do not know if I even started correctly. Thanks so much : )

ln(y) = sin(x)

y = e[sup:1n21elp1]sin(x)[/sup:1n21elp1]

dy/dx = cos(x) * e[sup:1n21elp1]sin(x)[/sup:1n21elp1]