Y = a ^ x, what is after differentiation (a is a constant)

Y = a ^ x, what is after differentiation (a is a constant)

This is a power function
dy=axlnadx

Let f (x) be continuous on [0,1], greater than 0 in (0,1), and satisfy the differential equation XF '(x) = f (x) + (3 / 2) ax & # 178; (a is a constant)
Let f (x) be continuous on [0,1], f (x) > 0 in (0,1), and satisfy the differential equation XF '(x) = f (x) + (3 / 2) ax & # 178; (a is a constant). Moreover, the area of the figure represented by the curves y = f (x) and x = 1, y = 0 is 2. Find out the function f (x), and ask what the value of a is. When the figure rotates around the X axis, the volume of the body of revolution is the smallest
I read the answers in the book, and I also find out the scope of A. I hope someone can find out the scope of a in detail, so that f (x) > 0 holds!

XY '= y + (3 / 2) ax & # - 178; the general solution of differential equation y = (3 / 2) ax & # - 178; + CXF (x) and x = 1, y = 0 form a graph area of 2 ∫ (0 → 1) YDX = 2 ∫ - A / 2 + C / 2 = 2, C = 4-af (x) = (3 / 2) ax & # - 178; + (4-A) XF (x) = (3 / 2) ax & # - 178; + (4-A) x has f (x) > 0A = 0, f (x) = 0 in X ∈ (0,1)

If a function is not differentiable at a certain point, it must not be differentiable. If it is not differentiable at a certain point, there must be no tangent,

It is correct that differentiability is not differentiable because differentiability is a necessary and sufficient condition
For example, if the tangent is perpendicular to the x-axis, then the slope of the tangent is infinite and does not exist. That is to say, there is a tangent at the point, but it is not differentiable

Why is the maximum static friction greater than the sliding friction of an object?

If the object is static on the table, then the rough part under the block will be embedded in the rough part above the table, just like the gear bite. In this way, to make the block move, there must be a part of force to overcome the downward component of gravity along the inclined plane

x/2+x/6+x/12+x/20+x/30+x/42=1
Such as problem solving, equation solving, online, etc

x/2+x/6+x/12+x/20+x/30+x/42=1
36x/42=1
x=7/6

A and B vehicles run from ab at the same time. A vehicle runs 60 km per hour, B vehicle runs 45 km per hour. The two vehicles meet at 30 km from the midpoint to find the distance between AB and ab

30 × 2 = 60 km 60 ÷ [60-45] = 4:4 × [60 + 45] = 420 km

（a+b）2+2（a+b）+1

Let a + B = t, then (a + b) 2 + 2 (a + b) + 1, = T2 + 2T + 1, = (T + 1) 2, that is, (a + b) 2 + 2 (a + b) + 1 = (a + B + 1) 2

Mathematics (1 + 3 + 5 + 7 +... + 999) - (2 + 4 + 6 +... + 998)

(1+3+5+7+...+999)-(2+4+6+...+998)
=1+（3-2）+(5-4)+(7-6)+.+(999-998)
=1+449×1
=500

The weight of three steel plates is 612kg. The weight of the first steel plate is three times that of the second steel plate, and the weight of the second steel plate is two times that of the third steel plate?

Let the third block be XKG, then there are:
2x+x+3*2x=612；
9x=612；
x=68；
The first one weighs 408kg, the second 136 kg and the third 68 kg

Given 10 ^ a = 3, 10 ^ B = 5, find the value of 10 ^ 3A + 2B

3^3*5*5=27*25=675