find an equation of the tangent line to the curve at the point (0, 0). y = sin(7 x)+(sin(7 x))^2 y(x) =
find an equation of the tangent line to the curve at the point (0, 0). y = sin(7 x)+(sin(7 x))^2 y(x) =
Answer
Explanation:
Step1: Find the derivative of $y$
Use the chain - rule. The derivative of $\sin(u)$ with respect to $x$ is $\cos(u)\cdot u'$. Let $u = 7x$. The derivative of $\sin(7x)$ is $7\cos(7x)$. For $(\sin(7x))^{2}$, let $v=\sin(7x)$, then by the chain - rule $\frac{d}{dx}(v^{2}) = 2v\cdot v'=2\sin(7x)\cdot7\cos(7x)$. So $y'=7\cos(7x)+14\sin(7x)\cos(7x)$.
Step2: Evaluate the derivative at $x = 0$
Substitute $x = 0$ into $y'$. When $x = 0$, $\cos(7x)=\cos(0)=1$ and $\sin(7x)=\sin(0)=0$. $y'(0)=7\cos(0)+14\sin(0)\cos(0)=7\times1 + 14\times0\times1=7$.
Step3: Use the point - slope form of a line
The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(0,0)$ and $m = y'(0)=7$. Substituting these values, we get $y-0 = 7(x - 0)$.
Answer:
$y = 7x$