how to prove that derivative of cos(x) is -sin(x)

Question : Prove that derivative of cos(x) is -sin(x)Solution :Let\[f\left(x\right)=\sin{\left(x\right)}\] By the definition of the derivative of a function\[\frac{df\left(x\right)}{dx}=limit_{∆x→0}\frac{f(x+∆x)-f(x)}{x+∆x-x}\] Using the definition of f(x) \[=limit_{∆x→0}\frac{cos(x+∆x)-cos(x)}{∆x}\] By the formula cos(A+B) =cos(A)cos(B)-sin(A)sin(B) \[\frac{df\left(x\right)}{dx}=limit_{∆x→0}\frac{cos(x)cos(∆x)-sin(x)sin(∆x)-cos(x)}{∆x}\] \[=limit_{∆x→0}\frac{cos(x)(cos∆x-1) -sin(x)sin(∆x)}{∆x} \] \[=cos\left(x\right)limit_{∆x→0}(\frac{cos(∆x)-1)}{ ∆x}-sin(x) limit_{∆x→0}\frac{sin(∆x)}{∆x}\]Since By L’Hospital rule…

This Question has been answered. 

Please Subscribe to See Answer or To Get Homework Help