How to find derivative of integral of u+1/u-2 from 4x to 9x?
Pretty sure it's: 9*(9x+1)/(9x-2) - 4*(4x+1)/(4x-2) This is because there is some F(x) that the integral is. When you evaluate at u = 4x, u = 9x and then differentiate you get f(9x) * d/dx (9x) - f(4x) * d/dx (4x). That's a chain rule issue. 9*(9x+1)/(9x-2) - 4*(4x+1)/(4x-2) = [9*(9x+1)(4x - 2) - 4(4x+1)(9x - 2)] / [(9x - 2)(4x - 2)] [9*(36x^2 - 14x - 2) - 4(36x^2 + x - 2)] / [(9x - 2)(4x - 2)] = (180x^2 - 130x - 10) / [(9x - 2)(4x - 2)] = (90x^2 - 65x - 5) / [(9x - 2)(2x - 1)] = 5*(18x^2 - 13x - 1) / [(9x - 2)(2x - 1)]
Find the derivative of g(x)=integral (u+5)/(u-3) du from 3x to 6x?
g(x) = integral(3x to 6x) (u+5)/(u-3) du = integral(3x to 0) (u+5)/(u-3) du + integral(0 to 6x) (u+5)/(u-3) du = - integral(0 to 3x) (u+5)/(u-3) du + integral(0 to 6x) (u+5)/(u-3) du So, applying the Fundamental Theorem of Calculus with the Chain Rule yields g'(x) = -[((3x) + 5)/((3x) - 3)] * 3 + [((6x) + 5)/((6x) - 3)] * 6. I hope this helps!
How do you find the derivative of the integral? Please help!!!?
Let f(u) = (u²-1)/(u²+1) Let F(u) = ∫ (u²-1)/(u²+1) du ----> F'(u) = f(u) g(x) = ∫₆ₓ⁷ˣ (u²-1)/(u²+1) du g(x) = F(7x) - F(6x) g'(x) = (F(7x))' - (F(6x))' . . . . . use chain rule: ....... = F'(7x) * 7 - F'(6x) * 6 ....... = 7 f(7x) - 6 f(6x) ....... = 7 ((7x)²-1)/((7x)²+1) - 6 ((6x)²-1)/((6x)²+1) ....... = 7 (49x²-1)/(49x²+1) - 6 (36x²-1)/(36x²+1) g'(x) = (1764x⁴ + 169x² -1) / ((49x² + 1)(36x² + 1)) ----------------------------------- Or, we could have rewritten original function as f(u) = (u²-1)/(u²+1) = (u²+1-2)/(u²+1) = 1 - 2/(u²+1) giving: g'(x) = 7 f(7x) - 6 f(6x) ....... = 7 (1 - 2/((7x)²+1)) - 6 (1 - 2/((6x)²+1)) ....... = 1 - 14/(49x²+1) + 12/(36x²+1) ----------------------------------- Note that the following are all equivalent: g'(x) = 7 (49x²-1)/(49x²+1) - 6 (36x²-1)/(36x²+1) g'(x) = (1764x⁴ + 169x² -1) / ((49x² + 1)(36x² + 1)) g'(x) = 1 - 14/(49x²+1) + 12/(36x²+1) .
Find the derivative of f(x)=xe^(4x)?
u gotta treat this as the derivative of the product of two functions... u look at xe^(4x) and ur prob like "wow wtfff"... but note that ur multiplying x by e^(4x)...now those are things that can be easily differentiated... now if ur taking the derivative of a product...u gotta follow the product rule... say ur taking the derivative of g(x)h(x), it turns out that the derivative of that is h ' (x) g (x) + h (x) g ' (x) where h ' (x) is the derivative of h(x) and so forth... so in this case, make f(x) = h(x)g(x) where h(x) = x and g(x) = e^(4x) now to get it outta the way, lets find out h ' (x) and g ' (x) h ' (x) = d/dx (x) = 1 (no brainer) g ' (x) = d/dx (e^(4x))...ok chain rule here...the rule goes that the derivative of e^(x) = e^(x)dx... so in this case its d/dx(4x) x e^(4x) = 4e^(4x) so combining them up, u get... h (x) g ' (x) + h ' (x) g (x) = x*4e^(4x) + 1*e^(4x) = 4xe^(4x) + e^(4x) ******= e^(4x) (4x + 1) ***** THAT'S THE ANSWER now to get the critical points, just set the above to equal 0... e^(4x) (4x+1) = 0 when does it equal 0?? well ur multiplying two things...if EITHER one of them is 0, the whole goshdarn thing is gonna be 0... so, the first part is e^(4x)...an exponent NEVER equals 0, unless its 0 as the base (i.e. 0^7 = 0 and always will be 0 lol)...so that part will nvr be 0 here... how about 4x + 1??? that can be 0... 4x + 1 = 0 4x = -1 x = -0.25 hope this helps -The Man
Integral question-find the derivative of the function?
The trick with these questions: when taking the derivative of an integral with functions of x at the bounds, do the following steps: Plug in the upper bound, use the chain rule, minus, plug in the lower bound, use the chain rule. There is no integration involved ... just the chain rule. Let us demonstrate. g(x) = Integral (6x to 9x, (u^2 - 1)/(u^2 + 1) du ) By the fundamental theorem of calculus, g'(x) = ( [9x]^2 - 1 ) / ( [9x]^2 + 1 ) {9} - ([6x]^2 - 1) / ([6x]^2 + 1) {6} I used the { } brackets to represent the use of the chain rule, since the derivative of 9x is 9, and the derivative of 6x is 6. Also, note how I plugged the function in directly in place of u. This is precisely how the Fundamental Theorem of Calculus works. The unfortunate truth is that this topic and these types of questions only get covered for a brief moment, so when an 8 mark question like this appears on a test, it's a lot of marks lost. Let me assist you by giving you more examples: EG 1) Find f'(x). f(x) = Integral ( sin(x) to e^x, sqrt(t^3 + 3) dt ) f'(x) = sqrt( (e^x)^3 + 3) {e^x} - sqrt( (sin(x))^3 + 3) {sin(x)} 2) f(x) = Integral ( x^2 to tan(x), sin(t^2) dt ) f'(x) = sin( [tan(x)]^2 ) {sec^2(x)} - sin( (x^2)^2 ) {2x} 3) Don't be fooled by constants, either! f(x) = Integral ( e^(2x) to 5, 1/[t^3 + 1] dt ) f'(x) = 1/[5^3 + 1] {0} - 1/[ (e^(2x))^3 + 1] (e^(2x))(2) The 0 cancels out the first term, leaving us with f'(x) = 2e^(2x) / [ e^(6x) + 1 ]
Find the derivative of the function.?
You drop the y because of the fact the slope of 7x is 7 The operation is addition so which you upload the derivatives, so which you're able to locate the by-manufactured from 7y and upload it to the by-manufactured from others. Given a linear equation y=6x, what's the slope? what's a derivative? a derivative is the slope of a factor, however the function is LINEAR, so it rather is not proper what the factor is, its in basic terms the slope. it rather is additionally the rationalization in the back of the by-manufactured from constants as 0. y=6 The function is linear. Slope=0 spinoff=comparable element, 0