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aek
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I'm having difficulties in differentiating this question y = x . (x-1)^1/2 , i keep on getting answers different to the back of the book. If someone can help, thnx
whozum said:[tex] x\sqrt{x+1}' = \sqrt{x+1}+\frac{x}{2\sqrt{x+1}} [/tex]
[tex] f(x)=x, g(x) = \sqrt{x+1} [/tex]
[tex] f(x)g(x) ' = f'(x)g(x) + f(x)g'(x) [/tex]
Hurkyl said:Either you're wrong (in which case you need to show your work), or the book's wrong. Actually, both could be wrong... but you seem to accept neither possibility.
I know for a fact that i did the power rule correct but simplifying and getting neat is where i find trouble.
And i asked did anyone else get an answer similar to the book.
im pretty sure your position would of been put under review if someone else can spare soo much time on this forum
Hurkyl said:You're welcome to talk to the management. You could PM someone directly, or use the "report bad post" feature, if you prefer. (It's the red triangle icon on the left of my post)
aek said:Your all geniuses, CLAP FOR THE HANDICAPS.
aek said:[tex] \frac{2(x+1)+x\sqrt{x+1}}{2(x+1)} [/tex]
It took my 4 days without a proper answer on PhysicsForum.com, however, it took less than 12 hours on a different forum which i refuse you people to join. Your all geniuses, CLAP FOR THE HANDICAPS.
Differentiation troubleshooting is the process of identifying and solving problems encountered when taking the derivative of a mathematical function. It involves understanding the rules and techniques of differentiation and applying them to equations to find the derivative.
To solve for y = x * (x-1)^1/2, you can use the product rule of differentiation. First, rewrite the equation as y = x * (x-1)^0.5. Then, apply the product rule: d/dx (f(x) * g(x)) = f'(x) * g(x) + f(x) * g'(x). In this case, f(x) = x and g(x) = (x-1)^0.5. After taking the derivative of each term, you can simplify to get the final answer in terms of x.
Some common mistakes when differentiating equations include forgetting to use the chain rule, not applying the product or quotient rule correctly, and making arithmetic errors. It is important to carefully follow the rules of differentiation and double check your work for any mistakes.
No, the power rule can only be applied when the exponent is a constant. In this case, the exponent is a variable (x) and therefore, the product rule must be used.
You can check if your differentiation is correct by taking the derivative of the original function and comparing it to your solution. If they are equal, then your answer is correct. You can also use online calculators or graphing software to graph both the original function and its derivative and see if they match up.