Differentiate sin x - x cos x
x cos x
x sin x
-x cos x
-x sin x
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So Its Me Anonymous time for me to solve this:
We need to differentiate: π¦ = sin π₯ β π₯cosπ₯
Step 1: Differentiate Term by Term
Using the derivative rules:
Derivative of sinπ₯:
πy/ππ₯(sinβ‘π₯)=cosβ‘π₯
Derivative of π₯cosβ‘π₯ (Product Rule): The product rule states:
πy/ππ₯[π’π£]=π’β²π£+π’π£β²
Let π’ = π₯ and π£ = cosπ₯,
then: π’β² = πy/ππ₯ π₯=1
π£β² = πy/ππ₯ cosπ₯ = βsinπ₯
Applying the product rule:
πy/ππ₯(π₯cosβ‘π₯)=(1)(cosβ‘π₯)+(π₯)(βsinπ₯) = cosπ₯ β π₯sinπ₯
Step 2: Put Everything Together
ππ¦ππ₯=cosβ‘π₯β(cosβ‘π₯βπ₯sinβ‘π₯)
Distribute the negative sign:
ππ¦ππ₯ = cosβ‘π₯ β cosπ₯ + π₯sinπ₯
β
Final Answer:
ππ¦ππ₯=π₯sinπ₯

sinx when differentiated gives cosx
gor xcos x
let u=x and v=cos x
du/dx=1 dv/dx=-sinx (differentiating individually first)
using the product rule i.e
dy/dx=du/dx ΓV +dv/dx ΓU
dy/dx=1Γcos x+(-sinx Γ x)
dy/dx=cosx-xsinx
put back in equation
cosx-(cosx-xsinx)
cosx-cosx+xsinx
=xsinx


