# Derivatives: Logarithmic Function help

## Homework Statement

Find an equation of the tangent line to the curve
$$\displaystyle \Large y = (2 x^2+5 )\ln (4 x^2-3 )+7$$
when x = 1.

## Homework Equations

$$\displaystyle \Large y = f'(x_0)(x-x_0)+f(x_0).$$

## The Attempt at a Solution

the fact that the tangent line to the curve y = f(x) when x = a is given by
$$\displaystyle \Large y = f'(x_0)(x-x_0)+f(x_0).$$ and a=1

f(1) = 7
f'(1) = 48
tangent line => y=48x-41
right?

this is how i did it:
f(1) = 7 (just plug in the number in f(x)
for f'(1),
i did:
$$\displaystyle \Large dy/dx = 4x\ln (4 x^2-3 )+(1/(4 x^2-3 ) )* 8x * (2 x^2+5 )$$
and pluged in 1 and got 48.

then found the tangent line by doing y=mx+b ==> y=48x+b ==> 7=48*1+b ==> b= -41
y=48x-41

is everything correct so far?

Mark44
Mentor
Everything looks fine, but I didn't check all of your arithmetic. Your value for f(1) is correct and your derivative is correct, but I didn't confirm it for f'(1).

I get another number for f'(1), not 48.

I get another number for f'(1), not 48.

omg.. what a stupid mistake... bhahaha, its 56 actually..
kk i got it lol.
thanks!