Differentiate cos⁡(x2+1) using first principle of derivative?

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Emily Green · Accepted Answer

ddxcos⁡(x2+1)For this problem, we need to use chain rule, as well as the fact that the derivative ofcos⁡(u)=−sin⁡(u). Chain rule basically just states that you can first derive the outside function with respect to what is inside the function, and then multiply this by the derivative of what is inside the function.Formally,dydx=dydu⋅dudx,where u=x2+1.We first need to work out the derivative of the bit inside the cosine, namely 2x.Then, after having found the derivative of the cosine (a negative sine), we can just multiply it by 2x.=−sin⁡(x2+1)⋅2x

Juan Goodwin · Accepted Answer

f(x)=cos⁡(x2−1)We need to findlimh→0f(x+h)−f(x)h=limh→0ocs((x+h)2−1)−cos⁡(x2−1)hLet&#039;s focus on the expression whose limit we need.cos⁡((x2−1)+(2xh+h2))−cos⁡(x2−1)h=cos⁡(x2−1)cos⁡(2xh+h2)−sin⁡(x2−1)sin⁡(2xh+h2)−cos⁡(x2−1)h=cos⁡(x2−1)cos⁡(2xh+h2)−1h−sin⁡(x2−1)sin⁡(2xh+h2)h=cos⁡(x2−1)cos⁡(2xh+h2)−1h−sin⁡(x2−1)sin⁡(2xh+h2)h=cos⁡(x2−1)cos⁡(2xh+h2)−1h(2x+h)(2x+h)−sin⁡(x2−1)sin⁡(2xh+h2)h(2x+h)(2x+h)We will use the following limits:limh→0cos⁡(2xh+h2)−1h(2x+h)=limt→0cos⁡t−1t=0limh→0sin⁡(2xh+h2)h(2x+h)=limt→0sin⁡tt=1And limh→0(2x+h)=2xTo evaluate the limit:cos⁡(x2−1)(0)(2x)−sin⁡(x2−1)⋅(1)⋅(2x)=−2xsin⁡(x2−1)

Differentiate \(\displaystyle{\cos{{\left({x}^{{{2}}}+{1}\right)}}}\) using first principle of derivative?

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