Derivatives of exponential functions

banfill_89
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Homework Statement



find the equation of the tangent to the curve defined by y=e^x that is perpendicular to the line defined by 3x+y=1

Homework Equations



m= 1/3 (negative reciprocal of y=-3x+1)
derivative of e^x=e^x

The Attempt at a Solution



i keep getting stuck at 1/3=e^x ( i know there's a way to solve for x i just can't remember)

the answer is x-3y+(1+ln3)=0
 
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ln(e^x) = x. You can take the natural log of both sides of an equation.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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