Question

for fun question.
Determine the partial derivative of cos(y+z) with respect to z using the definition of the derivative

Answer 1

@Angle

Answer 2

@Vocaloid

Answer 3

Intuitively I believe it would just be -sin(y+z)

I just learned partials so I could be wrong ;;

Answer 4

@vocaloid wouldn't you also have to multiply it by the derivative of the inside? (with respect to z)

Answer 5

but to ask this with the definition of the derivative is a bit more difficult ^_^"

Answer 6

Ok so yes your intuitive answer is correct @Vocaloid but the real fun comes with determining this using the definition of the derivative

Answer 7

and @angle the derivative of the interior of cos(y+z) with respect to z is 1 so 1*-sin(y+z) is still just -sin(y+z)

Answer 8

the goal is to use this
\[\lim_{\Delta z \rightarrow 0}\frac{ f(x,y,z+\Delta z)-f(x,y,z) }{ \Delta z }\]

Answer 9

\(\dfrac{\partial [\cos (y + z)]}{\partial z} := \lim\limits_{\delta z \to 0}\dfrac{cos (y + z + \delta z) - \cos (y + z)}{\delta z}\)

\(= \lim\limits_{\delta z \to 0}\dfrac{\cos (y + z) \cos \delta z - \sin (y + z) \sin \delta z - \cos (y + z)}{\delta z}\)

Re-arrange:

\(= \lim\limits_{\delta z \to 0} \left( \cos (y + z) \dfrac{\cos \delta z - 1}{\delta z} - \sin (y + z) \dfrac{\sin \delta z}{\delta z} \right)\)

Limit of sum is sum of limits:

\(= \cos (y + z) \lim\limits_{\delta z \to 0}\dfrac{\cos \delta z - 1}{\delta z} - \sin (y + z) \lim\limits_{\delta z \to 0} \dfrac{\sin \delta z}{\delta z} = \triangle\)

Both these trig limits are well known:

\( \left. \dfrac{\cos \delta z - 1}{\delta z} \right|_{\delta z \to 0} = 0\)

\( \left. \dfrac{\sin \delta z }{\delta z} \right|_{\delta z \to 0} = 1\)


So:

\(\triangle = \cos (y + z) \times 0 - \sin (y + z) \times 1\)

Link to the common trig limit + proofs:
http://tutorial.math.lamar.edu/Classes/CalcI/ProofTrigDeriv.aspx