i have two questions, for both i need to find F'(x) of the function.
i think both are using the product rule but not 100% certain. All help much appreciated.
1 f(x)= ln(x^3-1/3x^2)
2 f(x)=e^cosx *sin(ln x)
1.
$\displaystyle f(x)= \ln \left( \frac{x^3-1}{3x^2}\right)$
Using the log rules: $\displaystyle \ln \left(\frac{a}{b}\right) = \ln (a) - \ln(b)$
$\displaystyle f(x)= \ln \left( x^3-1\right) - \ln (3x^2)$
$\displaystyle \ln$ is given if in a fraction, the denominator differentiated gives the numerator. $\displaystyle \ln x = \int \frac{1}{x}$
$\displaystyle f'(x) = \frac{3x^2}{x^3-1} - \frac{6x}{3x^2}$
2.
$\displaystyle f(x)=e^{\cos x} *\sin(\ln x)$
$\displaystyle u = e^{\cos x}, \ u' = -\sin x . e^{\cos x}, \ v= \sin(\ln x), \ v' = \frac{\cos(\ln (x))}{x}$
$\displaystyle \frac{\mathrm{d}(uv)}{\mathrm{d}x} = uv' + u'v$
$\displaystyle f'(x) = (e^{\cos x})\left(\frac{\cos(\ln (x))}{x}\right) + (-\sin x . e^{\cos x})(\sin(\ln x))$
Hi!ollieman
Chain rule should be used for sureand
I will solve one of the question
as an example.(you may do other one by urself,ask if you have any doubt in solving)
y=f(x)=[e^cosx]sin(ln(x))
differentiating we get
y'=[d/dx(e^cosx)]sin(ln(x))+[e^cosx]d/dx(sin(lnx))
y'=[d/dx(e^cosx)]sin(ln(x))+[e^cosx]cos(lnx)d/dx(lnx)
y'=[d/dx(e^cosx)]sin(ln(x))+[e^cosx]cos(lnx)(1/x)
we may solve [d/dx(e^cosx)] saperately as it is a power function
let A=e^cosx
lnA=cosx
1/A(dA/dx)=-sinx
dA/dx=-sinx*A=-e^cos(x)[sin(x)]
so finally
y'= -[e^cos(x)]sin(x)sin(lnx)+[e^cosx]cos(lnx)(1/x)