Find the equation of the tangent to $y = 2^{x^2}$ at $(1,2)$?

Question

I First Found the First Derivative and i got..

$\ln(2)x⋅2^{x^2+1}$

Then I don;'t know what to do.. the real answer is $y = (4\ln2)x+2-4\ln2$

Answer 1

I'm afraid your formula contains a typo. The actual derivative of $f(x) = {2}^{{x}^{2}}$ is:

$$f'(x) = \mathrm{ln(2)}x\,{2}^{{x}^{2}+1}.$$

Thus the "1" is not a constant to be added to that expression; it is rather part of the exponent ($x^2 + 1$). Then,

$$f'(1) = 2.772588722239781 = 4\ln2.$$

At $(1,2)$, $2 = 4\ln2\cdot1 + c$. Therefore, $c = 2 - 4\ln2$. And finally:

$$y = (4\ln2)x + 2 - 4\ln2.$$

Which is in agreement with the answer you provided.

Answer 2

First you need to see that a give point is on the curve. Let $y=kx+l$ be are tangent line. Because given point is on the curve $k=y'(1)$. Let's find $y'$.
We know that $(a^x)'=ln(a)*a^x$. But in this case we have to use Chain rule to compute $y'$.

$$y'=2^{x^2}*ln(2)*2x$$ $$y'(1)=ln(2)*2=k$$ We know $x$,$y$ and $k$. And by just solving one equation for $l$. $$y=kx+l$$ $$y-kx=l$$ $$2-2ln(2)=l$$ And by that we know our tangent line.