Calculus Homework Help.

[chaheeto]

I'm stuck on these two Calculus problems. I don't understand how to solve them. What I've got so far is:

d(cos t)/dx = (-sin t)dt/dx

d(sin t)/dx = (cos t)dt/dx

d(ln t)/dx = (1/t)dt/dx.
When I apply those to the problems, I don't see how to make it evolve into the answer.
Any help is very much appreciated. Please show what steps I should take on solving this problem. Thank you very much.

#1. The derivative with respect to x of (ln x)/(sin x) isc

a. [(ln x)(cos x) - (1/x)(sin x)]/(ln x)^2
b. [(ln x)(sin x) + (1/x)(cos x)]/(cos x)^2
c. - [(ln x)(sin x) + (1/x)(cos x)]/(ln x)^2
d. [(1/x)(sin x) - (ln x)(cos x)]/(sin x)^2
e. none of these

#2. If f(x) = tan (ln x), then f '(x) =

a. cos (ln x)
b. -(1/x)[sin (ln x)]
c. [sec (ln x)]^2
d. -(1/x)[csc (ln x)]^2
e. sin (ln x)
f. (1/x)[cos (ln x)]
g. -[csc (ln x)]^2
h. (1/x)[sec (ln x)]^2
i. none of these

 

[galactus]

For #1, use the quotient rule, or rewrite as ln(x)csc(x) and use the product rule.

For #2. The chain rule comes in here. Derivative of inside times derivative of outside.

\(\displaystyle tan(ln(x))\)

Derivative of inside: \(\displaystyle \frac{d}{dx}[ln(x)]=\frac{1}{x}\)

Derivative of outside: \(\displaystyle \frac{d}{du}[tan(u)]=sec^{2}(u)\)

But, u=ln(x). So, we get \(\displaystyle sec^{2}(ln(x))\)

Multiply by the derivative of the inside: \(\displaystyle \frac{1}{x}sec^{2}(ln(x))=\frac{1}{xcos^{2}(ln(x))}\)

See?.

 

[BigGlenntheHeavy]

\(\displaystyle \frac{d[cos(t)]}{dx} \ = \ 0, \ \frac{d[cos(x)]}{dx} \ = \ -sin(x).\)