# Math Help - Calculus Problems

1. ## Calculus Problems

1. if $\frac{dy}{dx}=4y$ and if $y=4$ when $x=0$, then $y=$

2. $\int_0^{1/2} \frac{2x dx}{\sqrt{1-x^2}}=$

3. the point on the curve $2y=x^2$ nearest to $(4,1)$ is

4. if $tan(xy) = x$, then $\frac{dy}{dx}=$

2. Originally Posted by yamuda
1. if $\frac{dy}{dx}=4y$ and if $y=4$ when $x=0$, then $y=$

2. $\int_0^{1/2} \frac{2x dx}{\sqrt{1-x^2}}=$

3. the point on the curve $2y=x^2$ nearest to $(4,1)$ is

4. if $tan(xy) = x$, then $\frac{dy}{dx}=$
First one is an SDE so you go $\frac{dy}{y}=4dx$ integrating we get $ln|y|=4x+c$ then you get $y=Ce^{4x}$

2. The derivative of the quantity is present therefore $-\int_0^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}dx=-2\sqrt{1-x^2}|_0^{\frac{1}{2}}$

for the third one I think you can figure that out

4.differenating implicitly we get $sec^2(xy)\cdot{(xy'+y)}=1$ now solve for y'