Question

Can someone help me with this?

Answer 1

Question in attachment

Answer 2

hard to read answers but the derivative of the integral is the integrand. now use chain rule.
first suppose you had
\[\frac{d}{dx}\int_x^a t^{\frac{1}{2}}dt\] then this is the same as
\[\frac{d}{dx}-\int_a^x t{^\frac{1}{2}}dt\]
and that is just
\[-x^{\frac{1}{2}}\]

Answer 3

so that will be one term of your derivative. then we compute the next one
\[\frac{d}{dx}\int_a^{x^2} t^{\frac{1}{2}}dt\]

Answer 4

sorry should be x^4 up top. now it is a chain rule problem. the derivative of the integral is the integrand, but now you have to replace t by x^4 and also multiply by the derivative of x^4 via the chain rule. you get
\[(x^4)^{\frac{1}{2}}\times 4x^3\]
or
\[x^2\times 4x^3=4x^5\]

Answer 5

for a "final answer" of
\[4x^5-x^{\frac{1}{2}}\] or
\[4x^5-\sqrt{x}\]

Answer 6

Why -x^0.5 ?

Answer 7

Why -x^0.5 ?

Answer 8

the very first one computed. i can write it again

\[\frac{d}{dx}\int_a^x \sqrt{t}dt=\sqrt{x}\] since the derivative of the integral is the integrand yes?

Answer 9

so
\[\frac{d}{dx}\int_x^a \sqrt{t}dt = -\sqrt{x}\]

Answer 10

I still dont get it.How do u get the x^-0.5?