Find the derivative of the following: 1. h(x) = (3

Find the derivative of the following: 1. $h(x) = (3 - x^3)^2 e^{x^3} sin^2x$ 2. $g(x) = e^{cos^2(\frac{4x}{5} + 1)}$ 3. $f(x) = \frac{xe^{x \ln \sin x}}{sec x}$ 4. $y = \frac{4e^{\sqrt{x}}}{xcos x}$ 5. $w(x) = \frac{1 + csc 3x}{cot 3x}$ 6. $y = 5x^3 sin(x^{tan x})$ 7. $L(y) = e^{y^x} ln(ye^y)$ 8. $H(t) = (sin t)^t - cos t$ 9. $T(s) = \frac{sec \sqrt{s cos s^2}}{\sqrt{1 - cos s}}$ 10. $y = (3e^{\sqrt{x^2}} + e^{sin x})^x$

Transcript

00:01 Good day, the topic is about different derivatives.

00:03 According to chain rule, the derivative of a function f, that is also a function of g, is solved as the product of the derivative of function f and the derivative of function g.

00:13 Another useful technique in taking the derivative is the product rule, which states that the derivative of the product of the function u and v is equal to the u times the derivative of v plus v times the derivative of u.

00:27 We will apply these rules in evaluating the following functions or taking the derivatives of the following functions.

00:36 So let's start with the first one.

00:41 We note that the first expression can be rewritten such that we have 3 minus 2t times 1 plus sine 3t raised to negative 1.

00:54 Taking this is our u and this as our v then we can apply the product rule and get 3 minus 2 t times the derivative of b which is negative 1 times 1 plus sine 3 t raised to negative 2 multiplied by the derivative of 1 plus sine 3t which is first the derivative of 1 is 0 plus the derivative of sine 3 t is cosine 3 t multiplied by the derivative of 3 t which is so that's our u tv then add it with this time you copy the v 1 plus sign 3 t and then you take the derivative of your u so the derivative of 3 minus 2 t is 0 minus so this is your g prime let's simplify by factoring out the common factor which is 1 plus sine 3 t raised to negative 1 and that leaves us with negative 3 times 3 minus 2 t times 1 plus sine 3 t raised to negative 1 times cosine 3 t for the first term here and then minus 2 for the second term so this is the derivative of the function moving on to the next one we have r equals second square root of theta times tangent of 1 over theta we can rewrite this such that we have second of theta raised to the 1 half times tangent of theta raised to the negative 1 so again we can have this as our u and this is our v so applying the product rule gives us us first copy the u and then multiply it by the derivative of your v so the derivative of tangent is a second squared and then multiply by the derivative of theta to the negative one which is negative one theta raised to negative two so that's udv then this time let's copy the u rather to copy the and then multiply it with a derivative of your second theta raised to the one half.

03:44 So i'll be needing more room for that.

03:46 So let's put that here.

03:49 So again, that's plus tangent theta raise to negative 1 times second data raised to 1 half times tangent theta raised to 1 half...

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