Find the derivative of g(x)=cos(107x)
vbr{lr<5vvx><f(nn)/.bbn>
vbr<lr><5.12>(9.8)^3
.bbn=%ff
%ff=F(9x)
.bbn=%ff9x
.bbn=.133%tri
.bbn=.tri
t.ri=vbr
v=161.2
b=39.1
r=12.7
Sub in the inverse values of vbr
g ' (x) = -107 sin(107x)
To find the derivative of g(x) = cos(107x), we can use the chain rule.
The chain rule states that if we have a function g(x) = f(h(x)), where f(u) is differentiable with respect to u and h(x) is differentiable with respect to x, then the derivative of g(x) is given by:
g'(x) = f'(h(x)) * h'(x)
In this case, g(x) = cos(107x), so f(u) = cos(u) and h(x) = 107x.
Step 1: Find the derivative of f(u) = cos(u) with respect to u.
The derivative of cos(u) is -sin(u). So, f'(u) = -sin(u).
Step 2: Find the derivative of h(x) = 107x with respect to x.
The derivative of 107x is 107.
Step 3: Now, use the chain rule to find the derivative of g(x).
g'(x) = f'(h(x)) * h'(x)
= -sin(107x) * 107
So, the derivative of g(x) = cos(107x) is g'(x) = -107sin(107x).