Derieving Derivative #1

Calculus Level 2

The equation of the tangent to the curve y = b e − x / a y=be^ {-x/a} at the point where it cuts the y - axis is

b x + a y = 1 bx+ay=1 a x + b y = 1 ax+by=1 x / b + y / a = 1 x/b+y/a=1 x / a + y / b = 1 x/a+y/b=1

Parag Zode by Parag Zode , 24, India

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2 solutions

I have not done this concept yet in my school or either but i used a bit reasoning to solve it...For example, it says that the graph intersects the y-axis, so at y-axis the value of x=0. Putting that in the equation, we turn up with y=b. Now checking all the options only one is where x/a=0 (as x=0) and y/b=1(as y=b). Therefore, x/a+y/b=1.

CHEERS!!!

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The curve cuts the y-axis at x=0, so substituting 0 in the equation will give us the point of intersection (0,b). Getting the first derivative of the function will give us y"=(-b/a)(e^(-x/a)). Substituting x=0 in the equation of the derivative will give us the slope of the tangent line which is -b/a. Using the point-slope form, the equation of the line is y-b=(-b/a)(x-0). Simplifying a little bit will give us bx+ay=ab. Dividing both sides by ab will give us x/a-b/y=1 :D

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