Question

1. ((3t)6)'=2. (8x6+15x2-6x+10)'=3. (Sin x•tg x)'=​

Answer 1

1.((3t)6)'=(3t)'•(6)'=3•0=0

2. (8•6+15•2-6x+10)' =(48+30-6x+10)'=0+0-6+10=-6

3. (sinx•tgx)'=cox•tgx+sinx•(1/cos^2x)

*Если это производная, то, наверное, так

Answer 2

Sure, I'd be happy to help you with these derivatives!

1. Let's differentiate the expression ((3t)6) with respect to t:

   ((3t)6)' = 6 * (3t)^(6-1) * 3 = 18t^5.

2. Now, let's differentiate the expression (8x6+15x2-6x+10) with respect to x:

   (8x^6 + 15x^2 - 6x + 10)' = 6 * 8x^(6-1) + 2 * 15x^(2-1) - 6 = 48x^5 + 30x - 6.

3. Lastly, let's differentiate the expression (sin(x) * tan(x)) with respect to x:

   (sin(x) * tan(x))' = (cos(x) * tan(x)) + (sin(x) * sec^2(x)).

Here, sec(x) is the secant function, which is equal to 1/cos(x), and tan(x) is the tangent function.

So, the derivative of (sin(x) * tan(x)) is cos(x) * tan(x) + sin(x) * sec^2(x).

Please note that trigonometric identities are used in the differentiation of trigonometric functions.