# Thread: another differentiation of a function (bit harder!)

1. ## another differentiation of a function (bit harder!)

v = 3 sin (u) - 10tan (u) + 101n (u) + 2e^u

3cos (u) - 10/cos^2 (u) + 101/u + 2ue^u-1

maybe not right stuck with what to do on the -10tan and the last part

2. Originally Posted by JontyP
v = 3 sin (u) - 10tan (u) + 101n (u) + 2e^u

3cos (u) - 10/cos^2 (u) + 101/u + 2ue^u-1

maybe not right stuck with what to do on the -10tan and the last part
the derivative of $e^x$ is $e^x$

the derivative of a function of the form $a^x$, where $a>0$ is a constant and $x$ is our variable, is $a^x \ln a$

what is 101n (u) ?

what is the n? are there any powers here? (you should use ^ to indicate powers, so we know what you're talking about)

3. i did use ^ on the 2e ^ u is supposed to read 2e to the power of u

the n is just in the question it reads +101n (u) but im not 100% sure what to do with that

the e^u yeh that should be 2e^u after readin my notes again.

3cos (u) - 10/cos^2 (u) + 101/u + 2e^u

big question mark on the 101/u part however

4. Originally Posted by JontyP
i did use ^ on the 2e ^ u is supposed to read 2e to the power of u

the n is just in the question it reads +101n (u) but im not 100% sure what to do with that

the e^u yeh that should be 2e^u after readin my notes again.

3cos (u) - 10/cos^2 (u) + 101/u + 2e^u

big question mark on the 101/u part however
for 101n (u), are we to assume it looks like $101nu$? in that case, it's just using the power rule...

wait did you mean $\ln u$, as in the natural log of u? it is not a 1, it is a lower case L!

so, if you have $10 \ln u$ the derivative is $\frac {10}u$

and as i said, write $\sec^2 x$ instead of $\frac 1{\cos^2 x}$, it looks nicer

the rest are ok