How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #y = int (6+v^8)^3 dv# from 1 to cos(x)?

Question

How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #y = int (6+v^8)^3 dv#  from 1 to cos(x)?

Emily Ammerman · Accepted Answer

The answer is #-(6+cos^8(x))^3*sin(x)# Given a function #f# continuous on an interval #[a,b]# (with #a < b#) and a number #c\in [a,b]#, the Fundamental Theorem of Calculus ("Part 1"...actually, some books call it "Part 2") says that #d/dx(\int_{c}^{x}f(v)\ dv)=f(x)# for all #x\in [a,b]#. Thought of another way, the function #F# defined by the equation #F(x)=\int_{c}^{x}f(v)\ dv# is an antiderivative of #f# on the interval #[a,b]# (also, #F(c)=0#). For the problem at hand, let #F(x)=\int_{1}^{x}(6+v^8)^3\ dv# and let #g(x)=cos(x)# so that we must find #d/dx(F(g(x))#. This also requires the Chain Rule, which gives #d/dx(F(g(x)))=F'(g(x))*g'(x)#. But #F'(x)=(6+x^8)^3# and #g'(x)=-sin(x)#. The final answer is therefore: #d/dx(\int_{1}^{\cos(x)}(6+v^8)^3\ dv)=-(6+cos^8(x))^3*sin(x)#

Adam Adkins · Answer

undefined To find the derivative of the function $ y = \int_{1}^{\cos(x)} (6+v^8)^3 \, dv $, you can use Part 1 of the Fundamental Theorem of Calculus, which states that if $ F(x) $ is the integral of $ f(t) $ from a constant $ a $ to $ x $, then $ F'(x) = f(x) $.

Apply this theorem to the given function by first finding the antiderivative of $ (6+v^8)^3 $ with respect to $ v $. Let $ F(v) $ represent this antiderivative. Then, $ F'(\cos(x)) $ will be the derivative of the function with respect to $ x $.

Finally, find $ F'(\cos(x)) $ by differentiating $ F(v) $ with respect to $ v $ and then substituting $ \cos(x) $ for $ v $ in the resulting expression. This gives you the derivative of the function with respect to $ x $.