Question

Find the slope of the tangent line to the graph at the point (1,2). EQUATION BELOW

Answer 1

\[\sqrt{1+3^{x}}\]

Answer 2

I can get this far \[\frac{ \sqrt{1+3^{1+h}}-\sqrt{1+3^{1}} }{ h}\]

Answer 3

do you have to do it by 1st principles or can you just differentiate..?

Answer 4

I think just differentiate. I am new to calculus and was just told to solve it using the limit definition.

Answer 5

I can actually get it down to \[\frac{ \sqrt{1+3^{1+h}}-2 }{ h }\]

Answer 6

and that is for the limit as h approaches 0

Answer 7

I know i am suposed to transform the top so that I can cancel the h out of the denominator, but I have no clue as to how to get that square root to go away. Should i multiply by the conjugate on top and bottom of the whole numerator?

Answer 8

ok... so write it using index notation

\[y = ( 1 + 3^x)^{\frac{1}{2}}\]

the derivative of \[3^x ..... dy/dx = 3^x\]

so you will have

\[\frac{dy}{dx} = \frac{1}{2}(1 + 3^x)^{-\frac{1}{2}} \times 3^x\]

which simplifies to

\[\frac{dy}{dx} = \frac{3^x}{2\sqrt{1 + 3^x}}\]

Answer 9

I didn't really follow that..

Answer 10

I know how to use limit as h approaches 0 for f(x+h)-f(x)/h

Answer 11

and i can get it down to \[\frac{ -3+3^{1+h} }{ h(\sqrt{1+3^{1+h}}+2) }\]

Answer 12

but i'm stuck. I know i have to somehow cancel out that h on the bottom

Answer 13

well if you are going to use 1st principals you'll have

\[\lim_{x \rightarrow 0} \frac{\sqrt{1 + 3^{x + h}} - \sqrt{1 + 3^x}}{h}\]

you need to now multiply by the conjugate pair

\[\lim_{h \rightarrow 0} \frac{\sqrt{1 + 3^{x + h}} - \sqrt{1 + 3^x}}{h} \times \frac{\sqrt{1 + 3^{x + h}} + \sqrt{1 + 3^x}}{\sqrt{1 + 3^{x + h}} + \sqrt{1 + 3^h}}\]

Answer 14

oops the denominator should read 3^x

Answer 15

Ok, but i have to do it for the point (1,2) so shouldn't i also plug in 1 for where you have x's. if so then i posted above you my answer to what you said

Answer 16

but you can't use the point until you have the derivative.

so you would have

\[\lim_{h \rightarrow 0}\frac{1 + 3^{x + h} - 1 - 3^x}{h(\sqrt{1 + 3^{x + h}} + \sqrt{1 + 3^x})}\]

which will give

\[\lim_{h \rightarrow 0} \frac{3^x(3^h -1)}{h(\sqrt{1 + 3^{x + h}} + \sqrt{1 + 3^x})}\]

Answer 17

I should maybe add that the question says to "ESTIMATE the slope of the tangent line to 3 decimal places" if that makes any difference

Answer 18

Oh, i get what you're saying. But what after that? How can i get that h out of the exponent so i can cancel out the h on the bottom?

Answer 19

thats why just using the rules of differentiation make it so much easier...

I think the slope at x = 1 is m = 3/4

Answer 20

well i cant use those, but thanks anway. Guess i'll just ask my teacher on monday or something

Answer 21

ok.... it is an extremely tough question for 1st principles...

Answer 22

I think i may be over complicating it

Answer 23

because a later problem specifically mentions using the limit definition, but this one just says "Estimate to three decimal places the slope of the tangent line to the graph of y = f(x)
at the point (1, 2). Show the setup of your calculations and enough calculations to justify
your answer."

Answer 24

I'll just have to ask. Thanks anyway! ill still give ya a medal

Answer 25

nope problem... i just looked at wolframalpha and I'll attach the link....

http://www.wolframalpha.com/input/?i=differentiate+sqrt%7B1+%2B+3%5Ex%7D

Answer 26

thanks i couldn't get it into wolfram right, so thanks