Question
Verify directly that $F$ is an antiderivative of $f$$$F(x)=\sqrt{2 x^{2}-1} ; f(x)=\frac{2 x}{\sqrt{2 x^{2}-1}}$$
Step 1
F(x) = √(2x^2 - 1) Let u = 2x^2 - 1, then F(x) = √u and dF/dx = dF/du * du/dx. Show more…
Show all steps
Your feedback will help us improve your experience
Lucas Finney and 68 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Verify directly that $F$ is an antiderivative of $f$ $$F(x)=\frac{1}{3} x^{3}+2 x^{2}-x+2 ; f(x)=x^{2}+4 x-1$$
Integration
Antiderivatives and the Rules of Integration
Use differentiation to verify the antiderivative formula. $$f(u)=\sqrt{u} \text { and } u=x^{2}(x>0)$$
Differential Equations and Mathematical Modeling
Antidifferentiation by Substitution
In Exercises $1,2, \underline{3}$ and $\underline{4},$ verify directly that $F$ is an antiderivative of $f$. $$F(x)=\sqrt{2 x^{2}-1} ; f(x)=\frac{2 x}{\sqrt{2 x^{2}-1}}$$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD