I need a tiny bit favour from you guys. Given that [math]f(x)=\frac{x-1}{(x+2)^{50}}[/math]. The solution is [math]f'(x)=\frac{52 - 49x}{(x+2)^{-51}}[/math]. Am I correct?

 

Because the teacher had this solution: [math]f'(x)=\frac{52-49x}{(x+2)^{51}}[/math].

 

Who is right?

[math] \frac{df}{dx} = \frac{d}{dx} \left( (x-1) \cdot (x+2)^{-50} \right) = \left(\frac{d}{dx} (x-1) \right) \cdot (x+2)^{-50} +(x-1)\cdot \left( \frac{d}{dx} (x+2)^{-50} \right) [/math]

 

[math] = 1 (x+2)^{-50} + (x-1) \cdot -50 (x+2)^{-51} = \frac{x+2}{(x+2)^{51}} - 50 \frac{x-1}{(x+2)^{51}} = \frac{52 - 49x}{(x+2)^{51}} [/math]

 

Guess that point goes to your teacher (I did the calc because I thought you might both be wrong).

As a (very vague but hopefully helpfull) hint on derivatives: In polynomials, taking the derivative reduces the exponent by one. Here, you have something similar with the factor (x+2)^-50; the denominator. The answer does not necessarily have to look like the (x+2)^-51 which it does, but a jump from -50 to +51 (or +50 to -51 if it´s in the denominator) should be an alarm-sign that something can´t be correct with your solution.

Your teacher is correct. Check through your calculation - there will probably be a silly error there somewhere

Yeah thanks Atheist and dave, I got it. I know why it's positive 51, and not negative 51.

 

Meanwhile, I have another questions that I got confronted with.

 

[math]lim_{x \rightarrow 1} \frac{1}{(x-1)^2}[/math]

 

I assumed that the limit is not defined because there's a vertical asymptote at x=1. A teacher had this answer of [math]+ \infty[/math]. I thought this was wrong, because it would be correct if the limit goes [math]x \rightarrow 1^+[/math] or [math]x \rightarrow 1^-[/math].

 

What do you think?

How is that the derivative of [math]x^x[/math] is [math]x^x(lnx+1)[/math]??

 

Never mind I got it.

[math]lny = lnx^{x}[/math]

[math]lny={x}lnx[/math]

[math]\frac{y'}{y} = {x} \cdot \frac{1}{x} + (1)lnx[/math]

[math]y'=y(1+lnx)[/math]

[math]y'=x^x(1+lnx)[/math]

How do you figure [math]tan(arcsin(4/5))[/math] without a calculator?? Help me here please.

for the limit question look at the power in the denominator.

 

as for figuring the trig fuction, you can look at it like this.

 

because tan is sin divided by cosin the numerator becomes 4/5

 

now your left with the denominator, cos(arcsin(4/5)). consider the identity cos^2(x) +sin^2(x)=1

How is that the derivative of [math]x^x[/math] is [math]x^x(lnx+1)[/math]??

 

Never mind I got it.

[math]lny = lnx^{x}[/math]

[math]lny={x}lnx[/math]

[math]\frac{y'}{y} = {x} \cdot \frac{1}{x} + (1)lnx[/math]

[math]y'=y(1+lnx)[/math]

[math]y'=x^x(1+lnx)[/math]

 

A better way of doing it (that avoids implicit differentiation) is to rewrite [math]x^x = e^{x \log x}[/math], and simply use the chain rule. CPL.Luke's way is the best I can think of for your limit and trig problems.