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## Homework Statement

Find an equation of the tangent line to the curve

[tex]$\displaystyle \Large y = (2 x^2+5 )\ln (4 x^2-3 )+7$[/tex]

when x = 1.

## Homework Equations

[tex]$\displaystyle \Large y = f'(x_0)(x-x_0)+f(x_0).$[/tex]

## The Attempt at a Solution

the fact that the tangent line to the curve y = f(x) when x = a is given by

[tex]$\displaystyle \Large y = f'(x_0)(x-x_0)+f(x_0).$[/tex] 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:

[tex]$\displaystyle \Large dy/dx = 4x\ln (4 x^2-3 )+(1/(4 x^2-3 ) )* 8x * (2 x^2+5 )$[/tex]

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?