# Math Help - find derivative

1. ## find derivative

1. If f(x)=log_2[log_3(x)], find f'(4)

can someone check to see if my answers match yours?

f'(x)=1/[(xln3)]/[log_3(x)ln2]
=1/[xln3ln2log_3(x)]
f'(4)=1/[4ln3ln2log_3(4)]

2. Find the equation of the tangent to the curve defined by y=lnx-1 that is parallel to the straight line with equation 3x-6y-1=0.

M=0.5

y=lnx-1
y'=1/x
M_t=0.5
0.5=1/x
x=2

at x=2

y=lnx-1
y=ln2-1

y=mx+b
ln2-1=0.5(2)+b
ln2-1=b

therefore my answer is:

y=0.5x+ln2-1

2. Originally Posted by skeske1234
1. If f(x)=log_2[log_3(x)], find f'(4)

can someone check to see if my answers match yours?

f'(x)=1/[(xln3)]/[log_3(x)ln2]
=1/[xln3ln2log_3(x)]
f'(4)=1/[4ln3ln2log_3(4)]

2. Find the equation of the tangent to the curve defined by y=lnx-1 that is parallel to the straight line with equation 3x-6y-1=0.

M=0.5

y=lnx-1
y'=1/x
M_t=0.5
0.5=1/x
x=2

at x=2

y=lnx-1
y=ln2-1

y=mx+b
ln2-1=0.5(2)+b
ln2-1=b

therefore my answer is:

y=0.5x+ln2-1

first one is correct

$f'(x)=\frac{\frac{1}{x\ln (3)}}{\ln (2)\log _3 (x)}$

$f'(x)=\frac{1}{x\ln (2)\ln (3) \log _3 (x)}$

3. Originally Posted by Amer
first one is correct

$f'(x)=\frac{\frac{1}{x\ln (3)}}{\ln (2)\log _3 (x)}$

$f'(x)=\frac{1}{x\ln (2)\ln (3) \log _3 (x)}$
First one is correct.

Is the second one correct?
If not, could you explain. ty

4. the second one

$y-y_o = Slope(x-x_o)$

$x_o=2 ,y_o=ln(2)-1$

slope = 1/2

$y-(ln(2)-1)=.5(x-2)$

$y=.5(x) -1 +ln(2) -1$

$y=.5(x) +ln(2) -2$

you mistake when you find b value

b=ln(2)-2