Differential of this equation?

Question

I know this is the chain rule, but I'm rather rusty on my calculus techniques at the moment and am struggling:

y=x10^[ (344/825)x+(43/12) ] - c

where c is any real number.

Thank you.

Answer 1

y = x10^[ (344/825)x+(43/12) ] - c

Move c to the LHS and take ln,
ln( y+c) = ln x + [ (344/825)x+(43/12) ] (ln 10)

Differentiate both sides,
y'/(y+c) = 1/x + (344/825)( ln 10)

Substitute y+c and move it to the RHS,
y' = x10^[ (344/825)x+(43/12) ][1/x + (344/825)(ln 10)]

Simplify,
y' = 10^[ (344/825)x+(43/12) ][1 + (344x/825)( ln 10)]

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You can also use exponential function y = e^(lny) to start, which is little bit simpler.

Answer 2

y=x10^[ (344/825)x+(43/12) ] - c

Let a = 344/825 and b = 43/12, then
y = x * 10^(ax + b) - c

Use the product rule : d(uv)/dx = u * dv/dx + v * du/dx

Therefore, y' = x * d[10^(ax + b)]/dx + 1 * 10^(ax + b) + 0
noting that the derivative of the constant c is 0.

For d[10^(ax + b)]/dx, use the formula :
d(a^u)/dx = a^u * ln(a) * du/dx

Thus, y' = x * [10^(ax + b) * ln(10) * a] + 10^(ax + b)

Finally, y' = 10^(ax + b) * [ax * ln(10) + 1]

Substituting back for a and b gives :

y' = 10^[(344/825)x + 43/12] * [(344/825) * ln(10) * x + 1]

Answer 3

Given:y=x10^[(344/825)x+(43/12)]-c

step1:assign x as 'u' and 10^.... as 'v'.. ..and 344/34 as "a" and 43/12 as b to simplify matters......and forget about c as it gets canceled off under differentiation(real no. is a constant)
so y= x10^(ax+b)
2:dy/dx of uv= udv+vdu
This implies dy/dx= x.[dy/dx(10^(ax+b)]+10^(ax+b).dy/dx[x]....(1)

Now don't you feel the sum is easier???

Now you solve it by applying the BASIC formulas..
dy/dx[x]=1.....(2)
dy/dx(10^(ax+b)= [ax+b]10^[(ax+b)^-1].a........(3)

substitute the results of (2) and (3) in (1) and also the values of "a" and "b"... you will get the final answer...

Answer 4

Well, first off, c is useless. It gets cancelled out, so just cross it off the page. If you put this in correctly, it is more than just the chain rule. You need to use the product rule {(uv)'=v'u+u'v}, x being u and 10^... being v. Then, when you differentiate 10^... you need to use the chain rule. Any number to a power differentiated uses the rule (a^u)'=(a^u)(ln a)(du/dx). You should be able to do it on your own given those formulas.

If you can't figure it out on your own, and no one answers it, email me and I'll send you the answer.

Answer 5

Phew - I thought I passed maths A level. But it was 40+ year ago

Answer 6

I just finished an exam in this and i'm still clueless!

Answer 7

hahaha i wish i could help but im in algebra 1 AGAIN lol