Here, f=lnu, and u=sin (x). d/ (du)lnu=1/u d/dxsin (x)=cos (x) So (df (u))/dx=1/u*cos (x)=cos (x)/u. But remember, u=sin (x) So it becomes cos (x)/sin (x)=cot (x) Multiply this by the constant, 1/ln (10)*cot …

Explanation: First a couple of assumptions on this question: (i) you are looking for the first derivative # ( (dY)/dx)# (ii) #log# is natural log (#ln#)

The second can now be evaluated as Deltaxrarr0: f' (x)=1* (1+tan^2 (x))/ (cos (0) (1-tan (x)tan (0)) Since cos (0)=1 and tan (0)=0: f' (x)= (1+tan^2 (x))/ (1 (1-tan (x)*0)) f' (x)=1+tan^2 (x) Which is a common …

Explained below Given function f(x)= 1/x 1. This function is defined for all x on the number line, except at x=0. Hence the domain in interval form would be (-oo,0)U(0,oo) 2 &3.Vertical asymptote is x=0 or y- …

(0, 0) From Wikipedia: a critical point or stationary point of a differentiable function of a single real variable, f(x), is a value x_0 in the domain of f where its derivative is 0: f′(x_0) = 0 Thus, to find the …

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Let #g (x) = 1+ sin x# #color (blue) (g' (x) = cos x *1) =>color (blue) ( cosx) # Let's substitute into (1) to get #f' (x) = ( (1+ sinx)color (red) ( (-2sinx cos x ...

Explanation: First, recall the formula: #d/dx (tan^-1 (x)) = 1/ (1+x^2)# We will need chain rule for our derivative: #d/dx (tan^-1 (x+1/x))# #= d/ (d (x+1/x)) (tan^-1 ...

Jan 25, 2018 · #d/dx ( (ln (x))^3) = 3 (ln (x))^2*d/dx (ln (x))= (3 (ln (x))^2)/x#. Alternatively #y=u^3# and #u=ln (x)#: #dy/ (dx)=dy/ (du)* (du)/ (dx)# #dy/ (dx)=3u^2*1/x# and substituting #dy/ (dx)= (3 (ln …

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