- #1
Slimsta
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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?