hristo, where are you from?

I have a friend called Hristo and he's hungarian. I think that he comes from Hungary. Is it right?

]]>f(x)= x 3

4 x 5

¹(x)= 5x 3 + x xln3

And for this:

0.7x

y=(cosx)

I dunno if u have to calculate the derivative, but u can rewrite it in this way

(7/10)x ln(cosx)

f(x)= e

You can use this, and the chain rule as well. Normally, solving an implicit differentiation (where you don't have y by itself alone on one side of the equation) can lead to some complications later, but in this case, the derivative of ln(y) is 1 / y, so it works out pretty well.

]]>y = (cos x )^0.7x

lny = ln(cos x )^0.7x

lny = .7xln(cox)

give it a try! oh and happy holidays everyone!

]]>y = (cos x )^0.7x

thanks

]]>the answer in the book is

f' (x) = x^5 * 3^x *(5/x + ln 3)

]]>So, (x^5 * 3^x)' = 3^x(x^5)' + x^5(3^x)'

(x^5)' = 5*x^4, using the basic differentiation of powers rule.

3^x is slightly different, because x is the exponent, so this time the derivative is ln 3* 3^x.

So, your overall derivative is 5 * x^4 * 3^x + x^5 * ln 3 * 3^x

Factorise to make it neater: x^4*3^x(5+x*ln 3)

]]>how do i solve the following problem:

f (x) = x^5 * 3^x

f' (x) = ?

thanks in advance

]]>