# find the differential of each function

• Mar 2nd 2010, 12:14 AM
watchthesky30
find the differential of each function
The question is find the differential of each function: a) y=(t+tant)^5 and b) y=√(z+1/z)
a) dy/dt = 5(t+tan t)(1 +sec^2 x)
b) dy/dz= 0.5*(1/(z + 1/z)) * (1 - 1/z^2)
• Mar 2nd 2010, 01:48 AM
Prove It
Quote:

Originally Posted by watchthesky30
The question is find the differential of each function: a) y=(t+tant)^5 and b) y=√(z+1/z)
a) dy/dt = 5(t+tan t)(1 +sec^2 x)
b) dy/dz= 0.5*(1/(z + 1/z)) * (1 - 1/z^2)

I think you mean "find the derivative" of each function.

You will need to use the chain rule in each case.

a) $\displaystyle y = (t + \tan{t})^5$.

Let $\displaystyle u = t + \tan{t}$ so that $\displaystyle y = u^5$.

$\displaystyle \frac{du}{dt} = 1 + \sec^2{t} = 1 + \tan^2{t} + 1 = 2 + \tan^2{t}$.

$\displaystyle \frac{dy}{du} = 5u^4 = 5(t + \tan{t})^4$.

So $\displaystyle \frac{dy}{dt} = 5(2 + \tan{t})(t + \tan{t})^4$.

b) $\displaystyle y = \sqrt{z + \frac{1}{z}} = \left(z + z^{-1}\right)^{\frac{1}{2}}$.

Let $\displaystyle u = z + z^{-1}$ so that $\displaystyle y = u^{\frac{1}{2}}$.

$\displaystyle \frac{du}{dz} = 1 - z^{-2} = 1 - \frac{1}{z^2} = \frac{z^2 - 1}{z^2}$.

$\displaystyle \frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2\sqrt{z + \frac{1}{z}}}$.

Therefore $\displaystyle \frac{dy}{dz} = \frac{z^2 - 1}{2z^2\sqrt{z + \frac{1}{z}}}$.