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- Please don't format the response like this: Now find the derivative of the given function. h'x=dhxdx=ddxtan2x+5=sec22x+5·ddx2x+5 Apply the Chain rule=sec22x+5ddx2x+ddx5=sec22x+52·ddxx+5·ddx1=sec22x+521x1-1+50 Apply the Power law=sec22x+521+0=sec22x+52=2sec22x+5 Now find the second derivative of the given function. h''x=d2hxdx2=ddxdhxdx=ddx2sec22x+5=2·ddxsec22x+5=2ddxsec2x+52=22sec2x+52-1·ddxsec2x+5 Apply the Chain rule=22sec2x+51sec2x+5tan2x+5·ddx2x+5 Again apply the Chain rule=22sec22x+5tan2x+5ddx2x+ddx5=22sec22x+5tan2x+52ddxx+5ddx1=22sec22x+5tan2x+521x1-1+50 Apply the Power law=22sec22x+5tan2x+52=8sec22x+5tan2x+5 Therefore, h'x=2sec2(2x+5)h''x=8sec2(2x+5)tan(2x+5) It is difficult to read when I don't understand the problem and want to learnFind the derivative of the function. f(ss)equals=StartFraction StartRoot s EndRoot minus 3 Over StartRoot s EndRoot plus 5 EndFractions−3s+5 Question content area bottom Part 1 Choose the correct answer below. A.StartFraction 4 Over left parenthesis StartRoot s EndRoot minus 5 right parenthesis squared EndFraction StartFraction 4 Over left parenthesis StartRoot s EndRoot minus 5 right parenthesis squared EndFraction4s−52 B.StartFraction 4 Over StartRoot s EndRoot left parenthesis StartRoot s EndRoot minus 5 right parenthesis squared EndFraction StartFraction 4 Over StartRoot s EndRoot left parenthesis StartRoot s EndRoot minus 5 right parenthesis squared EndFraction4ss−52 C.StartFraction 4 Over StartRoot s EndRoot left parenthesis StartRoot s EndRoot plus 5 right parenthesis squared EndFraction StartFraction 4 Over StartRoot s EndRoot left parenthesis StartRoot s EndRoot plus 5 right parenthesis squared EndFraction4ss+52…Find the derivative of: f(z) = (3z+9) / (2-z) 15z + 3z/(2-z)^2 15z – 3z/(2-z)^2 15z/(2-z)^2 15/(2-z)^2 None of the above
- Evaluate the value of ‘a’ if the first derivative of m(x,y,z)=4-ax^2-2y^2, with respect to the independent variable x at the point (1, 2, 3), is -2. (m_x (1,2,3)= -2)Please don't reply with something like this. It's very hard to read and understand Explanation: y=6sinx at x=π6Find the value of the function at the given point: y0=fπ6=6sinπ6=6·12=3To find slope of the tangent line at x=x0 is the derivative of the function, evaluated at x=x0:m=f'(x0)f'(x)=ddx6sinx=6·ddxsinx=6cosxf'(x0)=6cos(x0)=6cosπ6=6·32=33So, m=f'(x0)=33Finally, the equation of the tangent line is y-y0=m(x-x0)We have m=33, y0=3, x0=π6⇒y-y0=m(x-x0)y-3=33x-π6y=33x-π6+3y=33x-33π6+3y=33x-3π2+3Hence, equation of tangent line is y=33x-3π2+3Is it true that every function y = ƒ(x) that is differentiable on [a, b] is itself the derivative of some function on [a, b] ? Give reasons for your answer.
- Give an example of a function f (x) that has one positive derivative on (−1,0) and a negative derivative on (0,1).The derivative of ƒ(x) = x2 is zero at x = 0, but ƒ is not a con-stant function. Doesn’t this contradict the corollary of the Mean Value Theorem that says that functions with zero derivatives are constant? Give reasons for your answer.Find the derivative of f(z)=5z(9ez−4). Include a multiplication sign between symbols. For example, a⋅π. f′(z)=