Using the Product Rule: \( y=uv \therefore y' = uv' - u'v \)
let \( u = 3x^{\frac{1}{2}} v = \sin (x) \) \( \therefore u' = \frac {3}{2} x^{-\frac {1}{2}} v' = \cos (x) \)
Substituting these values into the Product Rule: \( \therefore \frac {dy}{dx} = 3x^{\frac {1}{2}} \cos (x) - \frac{3}{2} x^{ -\frac {1}{2}} \sin (x) \)
Simplifying the equation into fractions leaves us with: \( \frac {dy}{dx} = 3 \sqrt x \cos(x) - \frac {3}{2\sqrt x} \sin(x) \)
As we cannot derive \( tan(\theta) \) with our current knowledge, but we know the \( \tan(\theta) = \frac {\sin(theta)}{\cos(\theta)} \) then \( g'(\theta) =\) the derivative of \( \frac {\sin(\theta)}{\cos(theta)} \)
Using The Quotient Rule when \( g = \frac {u}{v} \) then \( g' = \frac {vu' - uv'}{ v^2} \)
Let \( u = \sin(\theta) \) \(v = \cos(\theta) \) Then \( u' = \cos(\theta)\) \(v' = ^-\sin(\theta) \)
By substituting our values of \( u, v, u' \)and \(v'\) into The Quotient Rule
Example 7: The curve defined by \( y^3 + x^3 = 3xy \) is known as the Folium of Descartes, find the slope of the tangent to the curve at any point.
In order to find the slope of the tangent, we must differentiate the function. However, as we can not rearrange the function to the form y=y(x) we must use Implicit Differentiation.
As we are given the value of y to be \( 2^x \) we can substitute this value in:
\( frac {dy}{dx} = 2^x \ln(2) \)
Example 11: Differentiate \( y(x) = x^{\sin(x)} \)
As we have an exponent with x, we take the logarithms of both sides:
\( \log(y) = \log(x^{\sin(x)}) \)
Using our knowledge of Logarithmic Laws:
\( \ln(y) = \sin(x) \ln(x) \)
As we have two functions of x multiplied by each other, we will need to utilize The Product Rule: \( y=uv y' = v'u + u'v \) to derive the RHS of the equation.
Let \( u = sin(x) v = \ln(x) \) Let \( u'= cos(x) v' = \frac{1}{x} \)
Substituting these values into The Product Rule, yields:
Alright, well now we will have to figure out what \( \frac {du}{dx} \) is equal too so that the above equation will make sense. We are given at the start that:
\( y(x) = \arctan (u(x)) \)
\( \tan (y(x)) = u(x) \)
In order to find the value of \(\frac {du}{dx}\) we will need to differentiate both sides. We know from our Definitions List the derivative of \(\tan (y(x)) = sec^2 (y(x)) \)