What is the derivative of the function?

elton_fan
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hi all
i need a derivative of this function f(t)=100t/2t+15


thanks a lot
 
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You need to show us your attempt. Are you familiar with the quotient rule and so on?
 
FYI, you wrote

<br /> f(t)=\frac{100t}{2t}+15<br />

Is that what you meant? If you meant to write

<br /> f(t)=\frac{100t}{2t+15}<br />

then you need to use parentheses.
 
Hurkyl said:
FYI, you wrote

<br /> f(t)=\frac{100t}{2t}+15<br />

Is that what you meant? If you meant to write

<br /> f(t)=\frac{100t}{2t+15}<br />

then you need to use parentheses.

I'm almost certain he means the later, or the answer is pretty basic in fact it barely requires calculus.:smile:

<br /> f(t)=\frac{100t}{2t+15}<br />

Have you tried the quotient rule? Please show working?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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